661 lines
22 KiB
Plaintext
661 lines
22 KiB
Plaintext
# The Inverse of a Function
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In this section we will use these add-on packages:
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```julia
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using CalculusWithJulia
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using Plots
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```
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```julia; echo=false; results="hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "The Inverse of a Function",
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description = "Calculus with Julia: The Inverse of a Function",
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tags = ["CalculusWithJulia", "precalc", "the inverse of a function"],
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);
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nothing
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```
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----
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A (univariate) mathematical function relates or associates values of
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$x$ to values $y$ using the notation $y=f(x)$. A key point is a given
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$x$ is associated with just one $y$ value, though a given $y$ value
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may be associated with several different $x$ values. (Graphically, this is the vertical line test.)
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We may conceptualize such a relation in many ways: through an algebraic
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rule; through the graph of $f;$ through a description of what $f$
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does; or through a table of paired values, say. For the moment, let's
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consider a function as rule that takes in a value of $x$ and outputs a
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value $y$. If a rule is given defining the function, the computation
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of $y$ is straightforward. A different question is not so easy: for a
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given value $y$ what value - or *values* - of $x$ (if any) produce an output of
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$y$? That is, what $x$ value(s) satisfy $f(x)=y$?
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*If* for each $y$ in some set of values there is just one $x$ value, then this operation associates to each value $y$ a single value $x$, so it too is a function. When that is the case we call this an *inverse* function.
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Why is this useful? When available, it can help us solve equations. If we can write our equation as $f(x) = y$, then we can "solve" for $x$ through $x = g(y)$, where $g$ is this inverse function.
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Let's explore when we can "solve" for an inverse function.
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Consider the graph of the function $f(x) = 2^x$:
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```julia;hold=true
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f(x) = 2^x
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plot(f, 0, 4, legend=false)
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plot!([2,2,0], [0,f(2),f(2)])
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```
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The graph of a function is a representation of points $(x,f(x))$, so
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to *find* $f(c)$ from the graph, we begin on the $x$ axis at $c$, move
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vertically to the graph (the point $(c, f(c))$), and then move
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horizontally to the $y$ axis, intersecting it at $f(c)$. The figure
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shows this for $c=2$, from which we can read that $f(c)$ is about
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$4$. This is how an $x$ is associated to a single $y$.
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If we were to *reverse* the direction, starting at $f(c)$ on the $y$
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axis and then moving horizontally to the graph, and then vertically to
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the $x$-axis we end up at a value $c$ with the correct $f(c)$. This
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operation will form a function **if** the initial movement
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horizontally is guaranteed to find *no more than one* value on the graph. That
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is, to have an inverse function, there can not be two $x$ values
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corresponding to a given $y$ value. This observation is often
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visualized through the "horizontal line test" - the graph of a function with an
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inverse function can only intersect a horizontal line at most in one
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place.
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More formally, a function is called *one-to-one* *if* for any two $a
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\neq b$, it must be that $f(a) \neq f(b)$. Many functions are
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one-to-one, many are not. Familiar one-to-one functions are linear functions ($f(x)=a \cdot x + b$ with $a\neq 0$), odd powers of $x$ ($f(x)=x^{2k+1}$), and functions of the form $f(x)=x^{1/n}$ for $x \geq 0$. In contrast, all *even* functions are *not* one-to-one, as $f(x) = f(-x)$ for any nonzero $x$ in the domain of $f$.
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A class of functions that are guaranteed to be one-to-one are the
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*strictly* increasing functions (which satisfy $a < b$ implies $f(a) <
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f(b)$). Similarly, strictly decreasing functions are one-to-one. The
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term strictly *monotonic* is used to describe either strictly
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increasing or strictly decreasing. By the above observations,
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strictly monotonic function will have inverse functions.
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The function $2^x$, graphed above, is strictly increasing, so it will
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have an inverse function. That is we can solve for $x$ in an equation
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like $2^x = 9$ using the inverse function of $f(x) = 2^x$, provided we
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can identify the inverse function.
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## How to solve for an inverse function?
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If we know an inverse function exists, how can we find it?
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If our function is given by a graph, the process above describes how to find the inverse function.
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However, typically we have a rule describing our function. What is the
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process then? A simple example helps illustrate. The *linear*
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function $f(x) = 9/5\cdot x + 32$ is strictly increasing, hence has an
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inverse function. What should it be? Let's describe the action of $f$:
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it multiplies $x$ by $9/5$ and then adds $32$. To "invert" this we
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*first* invert the adding of $32$ by subtracting $32$, then we would
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"invert" multiplying by $9/5$ by *dividing* by $9/5$. Hence $g(x)=(x-32)/(9/5)$.
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We would generally simplify this, but let's not for now. If we
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view a function as a composition of many actions, then we find the
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inverse by composing the inverse of these actions in **reverse**
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order. The reverse order might seem confusing, but this is how we get
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dressed and undressed: to dress we put on socks and then shoes. To
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undress we take off the shoes and then take off the socks.
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When we solve algebraically for $x$ in $y=9/5 \cdot x + 32$ we do the same thing as we do verbally: we subtract $32$ from each side, and then divide by $9/5$ to isolate $x$:
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```math
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\begin{align}
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y &= 9/5 \cdot x + 32\\
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y - 32 &= 9/5 \cdot x\\
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(y-32) / (9/5) &= x.
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\end{align}
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```
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From this, we have the function $g(y) = (y-32) / (9/5)$ is the inverse function of $f(x) = 9/5\cdot x + 32$.
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*Usually* univariate functions are written with $x$ as the dummy variable, so it is typical to write $g(x) = (x-32) / (9/5)$ as the inverse function.
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*Usually* we use the name $f^{-1}$ for the inverse function of $f$, so this would be most often [seen](http://tinyurl.com/qypbueb) as $f^{-1}(x) = (x-32)/(9/5)$ or after simplification $f^{-1}(x) = (5/9) \cdot (x-32)$.
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!!! note
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The use of a negative exponent on the function name is *easily* confused for the notation for a reciprocal when it is used on a mathematical *expression*. An example might be the notation $(1/x)^{-1}$. As this is an expression this would simplify to $x$ and not the inverse of the *function* $f(x)=1/x$ (which is $f^{-1}(x) = 1/x$).
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##### Example
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Suppose a transformation of $x$ is given by $y = f(x) = (ax + b)/(cx+d)$. This function is invertible for most choices of the parameters. Find the inverse and describe it's domain.
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From the expression $y=f(x)$ we *algebraically* solve for $x$:
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```math
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\begin{align*}
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y &= \frac{ax +b}{cx+d}\\
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y \cdot (cx + d) &= ax + b\\
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ycx - ax &= b - yd\\
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(cy-a) \cdot x &= b - dy\\
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x &= -\frac{dy - b}{cy-a}.
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\end{align*}
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```
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We see that to solve for $x$ we need to divide by $cy-a$, so this expression can not be zero.
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So, using $x$ as the dummy variable, we have
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```math
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f^{-1}(x) = -\frac{dx - b}{cx-a},\quad cx-a \neq 0.
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```
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##### Example
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The function $f(x) = (x-1)^5 + 2$ is strictly increasing and so will have an inverse function. Find it.
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Again, we solve algebraically starting with $y=(x-1)^5 + 2$ and solving for $x$:
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```math
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\begin{align*}
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y &= (x-1)^5 + 2\\
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y - 2 &= (x-1)^5\\
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(y-2)^{1/5} &= x - 1\\
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(y-2)^{1/5} + 1 &= x.
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\end{align*}
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```
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We see that $f^{-1}(x) = 1 + (x - 2)^{1/5}$. The fact that the power $5$ is an odd power is important, as this ensures a unique (real) solution to the fifth root of a value, in the above $y-2$.
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##### Example
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The function $f(x) = x^x, x \geq 1/e$ is strictly
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increasing. However, trying to algebraically solve for an inverse
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function will quickly run into problems (without using specially
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defined functions). The existence of an inverse does not imply there
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will always be luck in trying to find a mathematical rule defining the
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inverse.
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## Functions which are not always invertible
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Consider the function $f(x) = x^2$. The graph - a parabola - is clearly not *monotonic*. Hence no inverse function exists. Yet, we can solve equations $y=x^2$ quite easily: $y=\sqrt{x}$ *or* $y=-\sqrt{x}$. We know the square root undoes the squaring, but we need to be a little more careful to say the square root is the inverse of the squaring function.
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The issue is there are generally *two* possible answers. To avoid this, we might choose to only take the *non-negative* answer. To make this all work as above, we restrict the domain of $f(x)$ and now consider the related function $f(x)=x^2, x \geq 0$. This is now a monotonic function, so will have an inverse function. This is clearly $f^{-1}(x) = \sqrt{x}$. (The ``\sqrt{x}`` being defined as the principle square root or the unique *non-negative* answer to ``u^2-x=0``.)
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The [inverse function theorem](https://en.wikipedia.org/wiki/Inverse_function_theorem) basically says that if $f$ is *locally* monotonic, then an inverse function will exist *locally*. By "local" we mean in a neighborhood of $c$.
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##### Example
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Consider the function $f(x) = (1+x^2)^{-1}$. This bell-shaped function is even (symmetric about $0$), so can not possibly be one-to-one. However, if the domain is restricted to $[0,\infty)$ it is. The restricted function is strictly decreasing and its inverse is found, as follows:
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```math
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\begin{align*}
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y &= \frac{1}{1 + x^2}\\
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1+x^2 &= \frac{1}{y}\\
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x^2 &= \frac{1}{y} - 1\\
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x &= \sqrt{(1-y)/y}, \quad 0 \leq y \leq 1.
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\end{align*}
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```
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Then $f^{-1}(x) = \sqrt{(1-x)/x}$ where $0 < x \leq 1$. The somewhat
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complicated restriction for the the domain coincides with the range of
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$f(x)$. We shall see next that this is no coincidence.
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## Formal properties of the inverse function
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Consider again the graph of a monotonic function, in this case $f(x) = x^2 + 2, x \geq 0$:
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```julia;hold=true
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f(x) = x^2 + 2
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plot(f, 0, 4, legend=false)
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plot!([2,2,0], [0,f(2),f(2)])
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```
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The graph is shown over the interval $(0,4)$, but the *domain* of $f(x)$ is all $x \geq 0$. The *range* of $f(x)$ is clearly $2 \leq x \leq \infty$.
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The lines layered on the plot show how to associate an $x$ value to a $y$ value or vice versa (as $f(x)$ is one-to-one). The domain then of the inverse function is all the $y$ values for which a corresponding $x$ value exists: this is clearly all values bigger or equal to $2$. The *range* of the inverse function can be seen to be all the images for the values of $y$, which would be all $x \geq 0$. This gives the relationship:
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> the *range* of $f(x)$ is the *domain* of $f^{-1}(x)$; furthermore the *domain* of $f(x)$ is the *range* for $f^{-1}(x)$;
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From this we can see if we start at $x$, apply $f$ we get $y$, if we then apply $f^{-1}$ we will get back to $x$ so we have:
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> For all $x$ in the domain of $f$: $f^{-1}(f(x)) = x$.
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Similarly, were we to start on the $y$ axis, we would see:
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> For all $x$ in the domain of $f^{-1}$: $f(f^{-1}(x)) = x$.
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In short $f^{-1} \circ f$ and $f \circ f^{-1}$ are both identity functions, though on possibly different domains.
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## The graph of the inverse function
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The graph of $f(x)$ is a representation of all values $(x,y)$ where $y=f(x)$. As the inverse flips around the role of $x$ and $y$ we have:
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> If $(x,y)$ is a point on the graph of $f(x)$, then $(y,x)$ will be a point on the graph of $f^{-1}(x)$.
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Let's see this in action. Take the function $2^x$. We can plot it by generating points to plot as follows:
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```julia; hold=true;
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f(x) = 2^x
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xs = range(0, 2, length=50)
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ys = f.(xs)
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plot(xs, ys, color=:blue, label="f")
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plot!(ys, xs, color=:red, label="f⁻¹") # the inverse
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```
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By flipping around the $x$ and $y$ values in the `plot!` command, we
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produce the graph of the inverse function - when viewed as a function
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of $x$. We can see that the domain of the inverse function (in red) is
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clearly different from that of the function (in blue).
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The inverse function graph can be viewed as a symmetry of the graph of
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the function. Flipping the graph for $f(x)$ around the line $y=x$ will
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produce the graph of the inverse function: Here we see for the graph
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of $f(x) = x^{1/3}$ and its inverse function:
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```julia; hold=true;
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f(x) = cbrt(x)
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xs = range(-2, 2, length=150)
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ys = f.(xs)
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plot(xs, ys, color=:blue, aspect_ratio=:equal, legend=false)
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plot!(ys, xs, color=:red)
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plot!(identity, color=:green, linestyle=:dash)
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x, y = 1/2, f(1/2)
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plot!([x,y], [y,x], color=:green, linestyle=:dot)
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```
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We drew a line connecting $(1/2, f(1/2))$ to $(f(1/2),1/2)$. We can
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see that it crosses the line $y=x$ perpendicularly, indicating that
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points are symmetric about this line. (The plotting argument
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`aspect_ratio=:equal` ensures that the $x$ and $y$ axes are on the
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same scale, so that this type of line will look perpendicular.)
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One consequence of this symmetry, is that if $f$ is strictly increasing, then so is its inverse.
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!!!note
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In the above we used `cbrt(x)` and not `x^(1/3)`. The latter usage assumes that $x \geq 0$ as it isn't guaranteed that for all real exponents the answer will be a real number. The `cbrt` function knows there will always be a real answer and provides it.
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### Lines
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The slope of $f(x) = 9/5 \cdot x + 32$ is clearly $9/5$ and the slope of the inverse function $f^{-1}(x) = 5/9 \cdot (x-32)$ is clearly $5/9$ - or the reciprocal. This makes sense, as the slope is the rise over the run, and by flipping the $x$ and $y$ values we merely flip over the rise and the run.
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Now consider the graph of the *tangent line* to a function. This concept will be better defined later, for now, it is a line "tangent" to the graph of $f(x)$ at a point $x=c$.
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For concreteness, we consider $f(x) = \sqrt{x}$ at $c=2$. The tangent line will have slope $1/(2\sqrt{2})$ and will go through the point $(2, f(2)$. We graph the function, its tangent line, and their inverses:
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```julia; hold=true;
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f(x) = sqrt(x)
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c = 2
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tl(x) = f(c) + 1/(2 * sqrt(2)) * (x - c)
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xs = range(0, 3, length=150)
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ys = f.(xs)
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zs = tl.(xs)
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plot(xs, ys, color=:blue, legend=false)
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plot!(xs, zs, color=:blue) # the tangent line
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plot!(ys, xs, color=:red) # the inverse function
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plot!(zs, xs, color=:red) # inverse of tangent line
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```
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What do we see? In blue, we can see the familiar square root graph along with a "tangent" line through the point $(2, f(2))$. The red graph of $f^{-1}(x) = x^2, x \geq 0$ is seen and, perhaps surprisingly, a tangent line. This is at the point $(f(2), 2)$. We know the slope of this tangent line is the reciprocal of the slope of the red tangent line. This gives this informal observation:
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> If the graph of $f(x)$ has a tangent line at $(c, f(c))$ with slope $m$, then the graph of $f^{-1}(x)$ will have a tangent line at $(f(c), c)$ with slope $1/m$.
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This is reminiscent of the formula for the slope of a perpendicular line, $-1/m$, but quite different, as this formula implies the two lines have either both positive slopes or both negative slopes, unlike the relationship in slopes between a line and a perpendicular line.
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The key here is that the shape of $f(x)$ near $x=c$ is somewhat related to the shape of $f^{-1}(x)$ at $f(c)$. In this case, if we use the tangent line as a fill in for how steep a function is, we see from the relationship that if $f(x)$ is "steep" at $x=c$, then $f^{-1}(x)$ will be "shallow" at $x=f(c)$.
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## Questions
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###### Question
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Is it possible that a function have two different inverses?
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```julia; hold=true; echo=false
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choices = [L"No, for all $x$ in the domain an an inverse, the value of any inverse will be the same, hence all inverse functions would be identical.",
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L"Yes, the function $f(x) = x^2, x \geq 0$ will have a different inverse than the same function $f(x) = x^2, x \leq 0$"]
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answ = 1
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radioq(choices, answ)
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```
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###### Question
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A function takes a value $x$ adds $1$, divides by $2$, and then subtracts $1$. Is the function "one-to-one"?
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```julia; hold=true; echo=false
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choices = [L"Yes, the function is the linear function $f(x)=(x+1)/2 + 1$ and so is monotonic.",
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L"No, the function is $1$ then $2$ then $1$, but not \"one-to-one\""
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]
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answ = 1
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radioq(choices, answ)
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```
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###### Question
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Is the function $f(x) = x^5 - x - 1$ one-to-one?
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```julia; hold=true; echo=false
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choices=[L"Yes, a graph over $(-100, 100)$ will show this.",
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L"No, a graph over $(-2,2)$ will show this."
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]
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answ = 2
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radioq(choices, answ)
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```
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###### Question
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A function is given by the table
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```verbatim
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x | y
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--------
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1 | 3
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2 | 4
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3 | 5
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4 | 3
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5 | 4
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6 | 5
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```
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Is the function one-to-one?
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```julia; hold=true; echo=false
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yesnoq(false)
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```
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###### Question
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A function is defined by its graph.
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```julia; hold=true; echo=false
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f(x) = x - sin(x)
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plot(f, 0, 6pi)
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```
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Over the domain shown, is the function one-to-one?
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```julia; hold=true; echo=false
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yesnoq(true)
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```
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###### Question
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Suppose $f(x) = x^{-1}$.
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What is $g(x) = (f(x))^{-1}$?
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```julia; hold=true; echo=false
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choices = ["``g(x) = x``", "``g(x) = x^{-1}``"]
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answ = 1
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radioq(choices, answ)
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```
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What is $g(x) = f^{-1}(x)$?
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```julia; hold=true; echo=false
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choices = ["``g(x) = x``", "``g(x) = x^{-1}``"]
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answ = 2
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radioq(choices, answ)
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```
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###### Question
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||
|
||
A function, $f$, is given by its graph:
|
||
|
||
```julia; echo=false;
|
||
k(x) = sin(pi/4 * x)
|
||
plot(k, -2, 2)
|
||
```
|
||
|
||
What is the value of $f(1)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
val = k(1)
|
||
numericq(val, 0.2)
|
||
```
|
||
|
||
What is the value of $f^{-1}(1)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
val = 2
|
||
numericq(val, 0.2)
|
||
```
|
||
|
||
What is the value of $(f(1))^{-1}$?
|
||
|
||
```julia; hold=true; echo=false
|
||
val = 1/k(1)
|
||
numericq(val, 0.2)
|
||
```
|
||
|
||
What is the value of $f^{-1}(1/2)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
val = 2/3
|
||
numericq(val, 0.2)
|
||
```
|
||
|
||
###### Question
|
||
|
||
A function is described as follows: for $x > 0$ it takes the square root, adds $1$ and divides by $2$.
|
||
|
||
What is the inverse of this function?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices=[
|
||
L"The function that multiplies by $2$, subtracts $1$ and then squares the value.",
|
||
L"The function that divides by $2$, adds $1$, and then takes the square root of the value.",
|
||
L"The function that takes square of the value, then subtracts $1$, and finally multiplies by $2$."
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
A function, $f$, is specified by a table:
|
||
|
||
```verbatim
|
||
x | y
|
||
-------
|
||
1 | 2
|
||
2 | 3
|
||
3 | 5
|
||
4 | 8
|
||
5 | 13
|
||
```
|
||
|
||
What is $f(3)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
numericq(5)
|
||
```
|
||
|
||
What is $f^{-1}(3)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
numericq(2)
|
||
```
|
||
|
||
What is $f(5)^{-1}$?
|
||
|
||
```julia; hold=true; echo=false
|
||
numericq(1/13)
|
||
```
|
||
|
||
What is $f^{-1}(5)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
numericq(3)
|
||
```
|
||
|
||
###### Question
|
||
|
||
Find the inverse function of $f(x) = (x^3 + 4)/5$.
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``f^{-1}(x) = (5y-4)^{1/3}``",
|
||
"``f^{-1}(x) = (5y-4)^3``",
|
||
"``f^{-1}(x) = 5/(x^3 + 4)``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
|
||
###### Question
|
||
|
||
Find the inverse function of $f(x) = x^\pi + e, x \geq 0$.
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
raw"``f^{-1}(x) = (x-e)^{1/\pi}``",
|
||
raw"``f^{-1}(x) = (x-\pi)^{e}``",
|
||
raw"``f^{-1}(x) = (x-e)^{\pi}``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
What is the *domain* of the inverse function for $f(x) = x^2 + 7, x \geq 0$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
raw"``[7, \infty)``",
|
||
raw"``(-\infty, \infty)``",
|
||
raw"``[0, \infty)``"]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
What is the *range* of the inverse function for $f(x) = x^2 + 7, x \geq 0$?
|
||
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
raw"``[7, \infty)``",
|
||
raw"``(-\infty, \infty)``",
|
||
raw"``[0, \infty)``"]
|
||
answ = 3
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
From the plot, are blue and red inverse functions?
|
||
|
||
```julia; hold=true; echo=false
|
||
f(x) = x^3
|
||
xs = range(0, 2, length=100)
|
||
ys = f.(xs)
|
||
plot(xs, ys, color=:blue, legend=false)
|
||
plot!(ys, xs, color=:red)
|
||
plot!(x->x, linestyle=:dash)
|
||
```
|
||
|
||
```julia; hold=true; echo=false
|
||
yesnoq(true)
|
||
```
|
||
|
||
|
||
From the plot, are blue and red inverse functions?
|
||
|
||
```julia; hold=true; echo=false
|
||
f(x) = x^3 - x - 1
|
||
xs = range(-2,2, length=100)
|
||
ys = f.(xs)
|
||
plot(xs, ys, color=:blue, legend=false)
|
||
plot!(-xs, -ys, color=:red)
|
||
plot!(x->x, linestyle=:dash)
|
||
```
|
||
|
||
```julia; hold=true; echo=false
|
||
yesnoq(false)
|
||
```
|
||
|
||
|
||
###### Question
|
||
|
||
The function $f(x) = (ax + b)/(cx + d)$ is known as a [Mobius](http://tinyurl.com/oemweyj) transformation and can be expressed as a composition of $4$ functions, $f_4 \circ f_3 \circ f_2 \circ f_1$:
|
||
|
||
* where $f_1(x) = x + d/c$ is a translation,
|
||
* where $f_2(x) = x^{-1}$ is inversion and reflection,
|
||
* where $f_3(x) = ((bc-ad)/c^2) \cdot x$ is scaling,
|
||
* and $f_4(x) = x + a/c$ is a translation.
|
||
|
||
For $x=10$, what is $f(10)$?
|
||
|
||
```julia; echo=false
|
||
𝒂,𝒃,𝒄,𝒅 = 1,2,3,5
|
||
f1(x) = x + 𝒅/𝒄; f2(x) = 1/x; f3(x) = (𝒃*𝒄-𝒂*𝒅)/𝒄^2 * x; f4(x)= x + 𝒂/𝒄
|
||
𝒇(x;a=𝒂,b=𝒃,c=𝒄,d=𝒅) = (a*x+b) / (c*x + d)
|
||
numericq(𝒇(10))
|
||
```
|
||
|
||
For $x=10$, what is $f_4(f_3(f_2(f_1(10))))$?
|
||
|
||
```julia; hold=true; echo=false
|
||
numericq(f4(f3(f2(f1(10)))))
|
||
```
|
||
|
||
The last two answers should be the same, why?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
L"As $f_4(f_3(f_2(f)_1(x))))=(f_4 \circ f_3 \circ f_2 \circ f_1)(x)$",
|
||
L"As $f_4(f_3(f_2(f_1(x))))=(f_1 \circ f_2 \circ f_3 \circ f_4)(x)$",
|
||
"As the latter is more complicated than the former."
|
||
]
|
||
answ=1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
|
||
Let $g_1$, $g_2$, $g_3$, and $g_4$ denote the inverse functions. Clearly, $g_1(x) = x- d/c$ and $g+4(x) = x - a/c$, as the inverse of adding a constant is subtracting the constant.
|
||
|
||
What is $g_2(x)=f_2^{-1}(x)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = ["``g_2(x) = x^{-1}``", "``g_2(x) = x``", "``g_2(x) = x -1``"]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
What is $g_3(x)=f_3^{-1}(x)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
raw"``c^2/(b\cdot c - a\cdot d) \cdot x``",
|
||
raw"``(b\cdot c-a\cdot d)/c^2 \cdot x``",
|
||
raw"``c^2 x``"]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
Given these, what is the value of $g_4(g_3(g_2(g_1(f_4(f_3(f_2(f_1(10))))))))$?
|
||
|
||
```julia; echo=false
|
||
g1(x) = x - 𝒅/𝒄; g2(x) = 1/x; g3(x) = 1/((𝒃*𝒄-𝒂*𝒅)/𝒄^2) *x; g4(x)= x - 𝒂/𝒄
|
||
val1 = g4(g3(g2(g1(f4(f3(f2(f1(10))))))))
|
||
numericq(val1)
|
||
```
|
||
|
||
What about the value of $g_1(g_2(g_3(g_4(f_4(f_3(f_2(f_1(10))))))))$?
|
||
|
||
```julia; hold=true; echo=false
|
||
val = g1(g2(g3(g4(f4(f3(f2(f1(10))))))))
|
||
numericq(val)
|
||
```
|