Answer:
[tex]\huge\boxed{f^{-1}(x)=\dfrac{1}{5}x-3}[/tex]
Step-by-step explanation:
[tex]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[/tex]
Answer:
[tex]f^{-1}(x)=\frac{x-12}{5}[/tex]
Step-by-step explanation:
Given:
[tex]f(x)=5x+12[/tex]
Replace f(x) with y:
[tex]y=5x+12[/tex]
Switch variables and solve for y:
[tex]x=5y+12[/tex]
[tex]x-12=5y[/tex]
[tex]\frac{x-12}{5}=y[/tex]
Simplify:
[tex]\frac{1}{5}x-\frac{12}{5}=y[/tex]
Replace y with f^-1(x):
[tex]f^{-1}(x)=\frac{1}{5}x-\frac{12}{5}[/tex]
Therefore, the inverse function for [tex]f(x)=5x+12[/tex] is [tex]f^{-1}(x)=\frac{1}{5}x-\frac{12}{5}[/tex].
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.
