Question

Let f(x) = 5x + 12. Find f^−1(x).

Answer 1

Answer:

$f^{-1}(x)=\frac{x-12}{5}$

Step-by-step explanation:

Given:

$f(x)=5x+12$

Replace f(x) with y:

$y=5x+12$

Switch variables and solve for y:

$x=5y+12$

$x-12=5y$

$\frac{x-12}{5}=y$

Simplify:

$\frac{1}{5}x-\frac{12}{5}=y$

Replace y with f^-1(x):

$f^{-1}(x)=\frac{1}{5}x-\frac{12}{5}$

Therefore, the inverse function for $f(x)=5x+12$ is $f^{-1}(x)=\frac{1}{5}x-\frac{12}{5}$.

Notice how in the graph attached of the inverse functions that they are symmetric about the line y=x. This is a crucial characteristic of inverse functions.

Answer 2

Answer:

$\huge\boxed{f^{-1}(x)=\dfrac{1}{5}x-3}$

Step-by-step explanation:

$f(x)=5x+15\to y=5x+15\\\\\text{exchange x to y and vice versa}\\\\x=5y+15\\\\\text{solve for}\ y\\\\5y+15=x\qquad|\text{subtract 15 from both sides}\\\\5y=x-15\qquad|\text{divide both sides by 5}\\\\y=\dfrac{1}{5}x-3$