Inverse Functions: Finding Inverse Functions Analytically
DEFINITION OF ONETOONE:
 Graph the function and apply the Horizontal Line Test to determine if the function is onetoone and thus has an inverse function.
 Take the function equation and replace f(x) by y.
 Switch the x and the y in the function equation and solve for y.
 Replace y by f ^{1}(x).
 Verify that f(x) and f ^{1}(x) are inverse functions.
º Show that f[f ^{1}(x)] = x.
º Show that the domain and range have been reversed.
Here is a review of the Horizontal Line Test.
HORIZONTAL LINE TEST:
A function f is onetoone and has an inverse function if and only if no horizontal line intersects the graph of f at more than one point.
Let's use these guidelines to determine the inverse of a function.
Step 1: Graph the function and apply the Horizontal Line Test to determine if the function is onetoone and thus has an inverse function. No horizontal line intersects the graph in more than one place. Thus the function is onetoone and has an inverse function. 


Step 2: Take the function equation and replace f(x) by y. 
$f\left(x\right)=\frac{3x}{2}$ $y=\frac{3x}{2}$ 

Step 3: Switch the x and y in the function equation and solve for y. 
$y=\frac{3x}{2}$ Original $x=\frac{3y}{2}$ Switch x and y 2x = 3  y Multiply by 2 y + 2x = 3 Add y y = 3  2x Subtract 2x 

Step 4: Replace y by f ^{1}(x). 
y = 3  2x f ^{1}(x) = 3  2x 

Step 5: Verify that f(x) and f ^{1}(x) are inverse functions. 1. Show that f[f ^{1}(x)] = x. 2. Show that f ^{1}[f(x)] = x. 3. Show that the domain and range have been reversed. 1. $f\left[{f}^{1}\left(x\right)\right]=f\left(32x\right)=\frac{3\left(32x\right)}{2}=\frac{33+2x}{2}=\frac{2x}{2}=x$ 2. ${f}^{1}\left[f\left(x\right)\right]={f}^{1}\left(\frac{3x}{2}\right)=32\left(\frac{3x}{2}\right)=3\left(3x\right)=33+x=x$ 3.
Verified, the inverse function of $f\left(x\right)=\frac{3x}{2}$ is ${f}^{1}\left(x\right)=32x$ 
Step 1: Graph the function and apply the Horizontal Line Test to determine if the function is onetoone and thus has an inverse function. No horizontal line intersects the graph in more than one place. Thus the function is onetoone and has an inverse function. 


Step 2: Take the function equation and replace f(x) by y. 
$f\left(x\right)=\sqrt{x+5}$ $y=\sqrt{x+5}$ 

Step 3: Switch the x and y in the function equation and solve for y. 
$y=\sqrt{x+5}$ Original $x=\sqrt{y+5}$ Switch x and y ${x}^{2}=y+5$ Square ${x}^{2}5=y$ Subtract 5 

Step 4: Replace y by f ^{1}(x). 
$y={x}^{2}5$ ${f}^{1}\left(x\right)={x}^{2}5$ 

Step 5: Verify that f(x) and f ^{1}(x) are inverse functions. 4. Show that f[f ^{1}(x)] = x. 5. Show that f ^{1}[f(x)] = x. 6. Show that the domain and range have been reversed. 1. $f\left[{f}^{1}\left(x\right)\right]=f\left({x}^{2}5\right)=\sqrt{\left({x}^{2}5\right)+5}=\sqrt{{x}^{2}}=x$ 2. ${f}^{1}\left[f\left(x\right)\right]={f}^{1}\left(\sqrt{x+5}\right)={\left(\sqrt{x+5}\right)}^{2}5=\left(x+5\right)5=x$ 3.
Verified, the inverse function of $f\left(x\right)=\sqrt{x+5}$ is ${f}^{1}\left(x\right)={x}^{2}5$ 
Related Links: Math algebra Conics: Classifying from General Equation Component Form and Magnitude Pre Calculus 
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