daviddoji/CalculusWithJuliaNotes.jl
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

use quarto, not Pluto to render pages

Browse Source
This commit is contained in:
jverzani
2022-07-24 16:38:24 -04:00
parent 93c993206a
commit 7b37ca828c
879 changed files with 793311 additions and 2678 deletions
Download Patch File Download Diff File Expand all files Collapse all files

66
CwJ/precalc/inversefunctions.jmd
View File

@@ -120,10 +120,9 @@ From this, we have the function $g(y) = (y-32) / (9/5)$ is the inverse function
*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 $f^{-1}(x) = (5/9) \cdot (x-32)$.
```julia; echo=false
note(L"""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 $f^{-1}(x) = 1/x$).
""")
```
!!! note
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 $f^{-1}(x) = 1/x$).
##### Example
@@ -277,10 +276,9 @@ 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.
```julia; echo=false
note(L"""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.
""")
```
!!!note
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.
### Lines
@@ -323,8 +321,8 @@ Is it possible that a function have two different inverses?
```julia; hold=true; echo=false
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.",
L"Yes, the function $f(x) = x^2, x \geq 0$ will have a different inverse than the same function $f(x) = x^2, x \leq 0$"]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -335,8 +333,8 @@ A function takes a value $x$ adds $1$, divides by $2$, and then subtracts $1$. I
choices = [L"Yes, the function is the linear function $f(x)=(x+1)/2 + 1$ and so is monotonic.",
L"No, the function is $1$ then $2$ then $1$, but not \"one-to-one\""
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -347,8 +345,8 @@ Is the function $f(x) = x^5 - x - 1$ one-to-one?
choices=[L"Yes, a graph over $(-100, 100)$ will show this.",
L"No, a graph over $(-2,2)$ will show this."
]
ans = 2
radioq(choices, ans)
answ = 2
radioq(choices, answ)
```
###### Question
@@ -396,16 +394,16 @@ What is $g(x) = (f(x))^{-1}$?
```julia; hold=true; echo=false
choices = ["``g(x) = x``", "``g(x) = x^{-1}``"]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
What is $g(x) = f^{-1}(x)$?
```julia; hold=true; echo=false
choices = ["``g(x) = x``", "``g(x) = x^{-1}``"]
ans = 2
radioq(choices, ans)
answ = 2
radioq(choices, answ)
```
###### Question
@@ -457,8 +455,8 @@ 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 $2$."
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -509,8 +507,8 @@ choices = [
"``f^{-1}(x) = (5y-4)^3``",
"``f^{-1}(x) = 5/(x^3 + 4)``"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
@@ -524,8 +522,8 @@ raw"``f^{-1}(x) = (x-e)^{1/\pi}``",
raw"``f^{-1}(x) = (x-\pi)^{e}``",
raw"``f^{-1}(x) = (x-e)^{\pi}``"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -536,8 +534,8 @@ choices = [
raw"``[7, \infty)``",
raw"``(-\infty, \infty)``",
raw"``[0, \infty)``"]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -550,8 +548,8 @@ choices = [
raw"``[7, \infty)``",
raw"``(-\infty, \infty)``",
raw"``[0, \infty)``"]
ans = 3
radioq(choices, ans)
answ = 3
radioq(choices, answ)
```
###### Question
@@ -620,8 +618,8 @@ choices = [
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."
]
ans=1
radioq(choices, ans)
answ=1
radioq(choices, answ)
```
@@ -631,8 +629,8 @@ 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``"]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
What is $g_3(x)=f_3^{-1}(x)$?
@@ -642,8 +640,8 @@ 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``"]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
Given these, what is the value of $g_4(g_3(g_2(g_1(f_4(f_3(f_2(f_1(10))))))))$?