Inverse Functions and Logarithms

Solve for $x$ in terms of $y$ .
Now you have $x=$ (a function of y) = $f^{-1}(y)$ .
If you wish to have $x$ be the independent variable (and if the variables $x$ and $y$ have no applicable meaning), simply exchange $x$ and $y$ to get $y=f^{-1}(x)$ .

Example 1: Find a formula for the inverse to

$f(x) = 2x+1$ .

Solution: Set

$y=2x+1$ . By definition,

$y = f(x)\quad \Longleftrightarrow\quad x = f^{-1}(y).$ Therefore, we have:

$\begin{eqnarray} 2x+1&=&y \cr 2x&=& y-1 \cr x &=& \frac{y-1}{2}. \end{eqnarray}$ So,

$x=f^{-1}(y) = \dfrac{y-1}{2}$ . You would leave the inverse in this form if, for example,

$y$ represents the cost of

$x$ widgets. In this case,

$y=f(x)$ tells us the cost,

$y$ , for any given number of widgets,

$x$ .

$x=f^{-1}(y)$ gives the number of widgets,

$x$ , for a given cost

$y$ . If

$x$ and

$y$ have no meaning, we may wish to rewrite the inverse of

$f$ with the independent variable

$x$ :

$f^{-1}(x) = \dfrac{x-1}{2}$ .

Warning: in general, $f^{-1}(x)$ is not the same as $f(x)^{-1}$ ! Inverses and reciprocals are different things.

Example 2: Find a formula for an inverse to $f(x) = 2x^2 + 5$ .

Solution: In this case, $f$ does not pass the horizontal line test, so $f$ does not have an inverse defined on the whole domain. Let's restrict the domain to $[0,\infty)$ , for example, so that now $f$ does pass the HLT. DO: Why does $f$ now pass the HLT? The range of $f$ is $[5,\infty)$ . Why? To find a formula for $f^{-1}$ , we write: $\begin{eqnarray} y&=&f(x) \Leftrightarrow x=f^{-1}(y) )\cr 2x^2 + 5 &=&y\cr x^2 &=& (y-5)/2 \cr x&=&\pm\sqrt{\frac{y-5}{2}}=f^{-1}(y). \end{eqnarray}$ But here, $x$ is not a function. So we must determine which square root we want. Since the range of $f^{-1}$ is the domain of $f$ , our answer must be in $[0,\infty)$ , so we want the positive root: $f^{-1}(y) = x=+ \sqrt{(y-5)/2}$ . This makes sense whenever $y \ge 5$ . The domain of $f^{-1}$ is $[5,\infty)$ , which is the range of $f$ . At this point, if $x$ and $y$ have no applicable meaning, you could switch $x$ and $y$ if needed to get $y=f^{-1}(x)=\sqrt{\frac{x-5}{2}}$ .

The Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

ff vs. f′f'

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

How to find a formula for an inverse function

$f$ vs. $f'$