Determine if a function is one-to-one given a verbal, numerical, graphical, or algebraic representation of the function and explain why a function must be one-to-one in order to have an inverse function
Identify the definition of the inverse function \(\inv{f}\) for one-to-one function \(f\) and describe the relationship between the domain/range of \(f\) and the domain/range of \(\inv{f}\)
Apply the cancellation properties of inverse functions
Find the inverse of a one-to-one function given a numerical, algebraic, or graphical representation
Apply the formula for the derivative of inverse functions
The table below gives the population \(P(t)\) of a bacterial culture, \(t\) hours after it is introduced to an agar-filled petri dish.
\(t\) hours
0
1
2
3
4
5
6
7
8
\(N=P(t)\) bacteria
150
165
182
200
220
243
267
294
324
The inverse function \(\inv{P}(N)\text{,}\) gives the time elapsed since a bacterial culture was introduced to an agar-filled petri dish, when the population is \(N\) bacteria. Use this information to fill out the table below.
A function \(f\) is said to be one-to-one if it never takes on the same value twice, or in notation, \(f(x_1) \neq f(x_2)\) when \(x_1\neq x_2\text{.}\)
Theorem6.1.3.Horizontal Line Test.
A function is one-to-one if and only if no horizontal line intersects its graph more than once.
Any even power function (in fact, any even function) is not one-to-one, as \(f(-a) = f(a)\) for any \(a\) in the domain of \(f\text{.}\)
Definition6.1.7.Inverse Function.
Let \(f\) be a one-to-one function with domain \(A\) and range \(B\text{.}\) The inverse function is notated \(\inv{f}\text{,}\) with domain \(B\) and range \(A\text{.}\) The inverse function is defined by the equation \(\inv{f}(y) = x\iff f(x) = y\) for any \(y\in B\text{.}\)
Domain and Range of Inverse Functions.
The domain of \(\inv{f}\) is the range of \(f\)
The range of \(\inv{f}\) is the domain of \(f\)
Notation Alert!
\(\inv{f}\) is a special notation to indicate the function inverse; you should not confuse this with the notation for the multiplicative inverse (reciprocal), such as \(\inv{x}\text{.}\) That is,
\(\inv{f}(x)\) denotes the inverse of a function
\(\inv{x}\) denotes the multiplicative inverse of a variable, i.e. \(\inv{x} = \dfrac{1}{x}\)
Example6.1.8.
Use the table below to answer the questions. If the answer does not exist, write DNE.
There is a graphical interpretation of algebraically finding an inverse:
There is a picture drawn in the pre-class video that I can’t replicate right now.
The idea is that when algebraically finding an inverse function, you are finding the equation of the function that results when you reflect \(y=f(x)\) across the line \(y=x\)
Subsection6.1.2Pre-Class Activities
Example6.1.11.
If \(f(x) = x^5 + x^3 + x\text{,}\) find \(\inv{f}(3)\) and \(f(\inv{f}(2))\text{.}\)
No; if \(f(x) = x^3\text{,}\) then its inverse is \(\inv{f}(x) = x^{1/3}\text{,}\) which is not differentiable at \(x=0\text{.}\)
Derivative of Inverses (at a Point).
If \(f\) is a one-to-one, differentiable function at \(x = a\) with inverse function \(\inv{f}\) and \(f'(\inv{f}(a))\neq 0\text{,}\) then the inverse function is differentiable at \(a\) and
We can choose two restricted domains: either \((-\infty,2]\) or \([2,\infty)\text{.}\) Generally, we’ll choose the positive portion of the domain for the restricted domain, so choose \([2,\infty)\text{.}\)
\(\inv{h}(x) = 2 + \sqrt{\dfrac{x+8}{2}}\) on \([2,\infty)\)
Suppose \(\inv{f}\) is the inverse function of a differentiable function \(f\) with \(f(4) = 5\) and \(f'(4) = \dfrac{2}{3}\text{.}\) Find \((\inv{f})'(5)\text{.}\)
Suppose \(\inv{f}\) is the inverse function of a differentiable function \(f\text{,}\) and let \(G(x) = \dfrac{1}{\inv{f}(x)}\text{.}\) If \(f(3) = 2\) and \(f'(3) = \dfrac{1}{9}\text{,}\) find \(G'(2)\text{.}\)
![A graph of the function \(h(x) = 2x^2-8x\) on the interval \([-2,6]\)](generated/latex-image/image-165.svg)
