edits

quarto/precalc/inversefunctions.qmd

@@ -15,10 +15,17 @@ plotly()
 ---
 
 
-A (univariate) mathematical function relates or associates values of $x$ to values $y$ using the notation $y=f(x)$. A key point is a given $x$ is associated with just one $y$ value, though a given $y$ value may be associated with several different $x$ values. (Graphically, this is the vertical line test.)
+A (univariate) mathematical function relates or associates values of $x$ to values $y$ using the notation $y=f(x)$. A key point is a given $x$ is associated with just one $y$ value, though a given $y$ value may be associated with several different $x$ values. (Graphically, this is the horizontal line test.)
 
 
-We may conceptualize such a relation in many ways: through an algebraic rule; through the graph of $f;$ through a description of what $f$ does; or through a table of paired values, say.  For the moment, let's consider a function as a rule that takes in a value of $x$ and outputs a value $y$. If a rule is given defining the function, the computation of $y$ is straightforward. A different question is not so easy: for a given value $y$ what value - or *values* - of $x$ (if any) produce an output of $y$? That is, what $x$ value(s) satisfy $f(x)=y$?
+We may conceptualize such a relation in many ways:
+
+* through an algebraic rule;
+* through the graph of $f$;
+* through a description of what $f$ does;
+* or through a table of paired values, say.
+
+For the moment, let's consider a function as a rule that takes in a value of $x$ and outputs a value $y$. If a rule is given defining the function, the computation of $y$ is straightforward. A different question is not so easy: for a given value $y$ what value--or *values*--of $x$ (if any) produce an output of $y$? That is, what $x$ value(s) satisfy $f(x)=y$?
 
 
 *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.
@@ -80,7 +87,7 @@ p
 The graph of a function is a representation of points $(x,f(x))$, so to *find* $y = f(c)$ from the graph, we begin on the $x$ axis at $c$, move vertically to the graph (the point $(c, f(c))$), and then move horizontally to the $y$ axis, intersecting it at $y = f(c)$. The figure shows this for $c=2$, from which we can read that $f(c)$ is about $4$. This is how an $x$ is associated to a single $y$.
 
 
-If we were to *reverse* the direction, starting at $y = f(d)$ on the $y$ axis and then moving horizontally to the graph, and then vertically to the $x$-axis we end up at a value $d$ with the correct $f(d)$. This allows solving for $x$ knowing $y$ in $y=f(x)$.
+If we were to *reverse* the direction, starting at $y = f(d)$ on the $y$ axis and then moving horizontally to the graph, and then vertically to the $x$-axis we end up at a value $d$ with the correct value of $f(d)$. This allows solving for $x$ knowing $y$ in $y=f(x)$.
 
 The operation described will form a function **if** the initial movement horizontally is guaranteed to find *no more than one* value on the graph. That is, to have an inverse function, there can not be two $x$ values corresponding to a given $y$ value. This observation is often visualized through the "horizontal line test" - the graph of a function with an inverse function can only intersect a horizontal line at most in one place.
 
@@ -195,7 +202,7 @@ In the section on the [intermediate value theorem](../limits/intermediate_value_
 ## Functions which are not always invertible
 
 
-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.
+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.
 
 
 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 $u^2-x=0$.)
@@ -231,8 +238,9 @@ Consider again the graph of a monotonic function, in this case $f(x) = x^2 + 2,
 ```{julia}
 #| hold: true
 f(x) = x^2 + 2
-plot(f, 0, 4, legend=false)
-plot!([2,2,0], [0,f(2),f(2)])
+plot(f, 0, 4; yticks=[2,4,8,16],
+     legend=false, framestyle=:origin)
+plot!([(2,0), (2, f(2)), (0, f(2))])
 ```
 
 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 $2 \leq y \leq \infty$.
@@ -241,24 +249,18 @@ The graph is shown over the interval $(0,4)$, but the *domain* of $f(x)$ is all
 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:
 
 
-> the *range* of $f(x)$ is the *domain* of $f^{-1}(x)$; furthermore the *domain* of $f(x)$ is the *range* for $f^{-1}(x)$;
-
-
+> * the *domain* of $f^{-1}(x)$ is the *range* of $f(x)$;
+> * the *range*  of $f^{-1}(x)$ is the *domain* of $f(x)$;
 
 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:
 
 
 > For all $x$ in the domain of $f$: $f^{-1}(f(x)) = x$.
 
-
-
 Similarly, were we to start on the $y$ axis, we would see:
 
-
 > For all $x$ in the domain of $f^{-1}$: $f(f^{-1}(x)) = x$.
 
-
-
 In short $f^{-1} \circ f$ and $f \circ f^{-1}$ are both identity functions, though on possibly different domains.
 
 
@@ -280,11 +282,12 @@ Let's see this in action. Take the function $2^x$. We can plot it by generating
 f(x) = 2^x
 xs = range(0, 2, length=50)
 ys = f.(xs)
-plot(xs, ys, color=:blue, label="f")
-plot!(ys, xs, color=:red, label="f⁻¹") # the inverse
+plot(xs, ys; color=:blue, label="f",
+     aspect_ratio=:equal, framestyle=:origin, xlims=(0,4))
+plot!(ys, xs; color=:red, label="f⁻¹") # the inverse
 ```
 
-By flipping around the $x$ and $y$ values in the `plot!` command, we produce the graph of the inverse function - when viewed as a function of $x$. We can see that the domain of the inverse function (in red) is clearly different from that of the function (in blue).
+By flipping around the $x$ and $y$ values in the `plot!` command, we produce the graph of the inverse function--when viewed as a function of $x$. We can see that the domain of the inverse function (in red) is clearly different from that of the function (in blue).
 
 
 The inverse function graph can be viewed as a symmetry of the graph of the function. Flipping the graph for $f(x)$ around the line $y=x$ will produce the graph of the inverse function: Here we see for the graph of $f(x) = x^{1/3}$ and its inverse function:
@@ -295,15 +298,16 @@ The inverse function graph can be viewed as a symmetry of the graph of the funct
 f(x) = cbrt(x)
 xs = range(-2, 2, length=150)
 ys = f.(xs)
-plot(xs, ys, color=:blue, aspect_ratio=:equal, legend=false)
-plot!(ys, xs, color=:red)
-plot!(identity, color=:green, linestyle=:dash)
-x, y = 1/2, f(1/2)
-plot!([x,y], [y,x], color=:green, linestyle=:dot)
+plot(xs, ys; color=:blue,
+     aspect_ratio=:equal, legend=false)
+plot!(ys, xs; line=(:red,))
+plot!(identity; line=(:green, :dash))
+x = 1/4
+y = f(x)
+plot!([(x,y), (y,x)]; line=(:green, :dot))
 ```
 
-We drew a line connecting $(1/2, f(1/2))$ to $(f(1/2),1/2)$. We can see that it crosses the line $y=x$ perpendicularly, indicating that points are symmetric about this line. (The plotting argument `aspect_ratio=:equal` ensures that the $x$ and $y$ axes are on the same scale, so that this type of line will look perpendicular.)
-
+We drew a line connecting $(1/4, f(1/4))$ to $(f(1/4),1/4)$. We can see that it crosses the line $y=x$ perpendicularly, indicating that points are symmetric about this line. (The plotting argument `aspect_ratio=:equal` ensures that the $x$ and $y$ axes are on the same scale, so that this type of line will look perpendicular.)
 
 One consequence of this symmetry, is that if $f$ is strictly increasing, then so is its inverse.
 
@@ -472,7 +476,7 @@ plotly()
 nothing
 ```
 
-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 $x=f(c)$.
+The key here is that the shape of $f(x)$ near $x=c$ is directly 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 $x=f(c)$.
 
 
 ## Questions