Question

Given f(x)=3x+12, find f(f^-1(0))

Answer 1

Answer:

0

Step-by-step explanation:

METHOD 1.

We know:

$f\bigg(f^{-1}(x)\bigg)=x$

Therefore

$f\bigg(f^{-1}(0)\bigg)=0$

METHOD 2.

$\text{Find}\ f^{-1}(x)$

$f(x)=3x+12\to y=3x+12$

exchange x to y and vice versa:

$x=3y+12$

solve for y:

$3y+12=x$            subtract 12 from both sides

$3y=x-12$        divide both sides by 3

$y=\dfrac{x-12}{3}\to f^{-1}(x)=\dfrac{x-12}{3}$

$f\bigg(f^{-1}(x)\bigg)\to$    replace x in f(x) with the expression$\dfrac{x-12}{3}$

$f\bigg(f^{-1}(x)\bigg)=3\cdot\dfrac{x-12}{3}+12=x-12+12=x$

$f\bigg(f^{-1}(0)\bigg)\to$ put x = 0 to $f\bigg(f^{-1}(x)\bigg)=x$

$f\bigg(f^{-1}(0)\bigg)=0$

METHOD 3.

$\text{Find}\ f^{-1}(x)$

$f^{-1}(x)=\dfrac{x-12}{3}$

Calculate the value of f ⁻¹(x) for x = 0:

$f^{-1}(0)=\dfrac{0-12}{3}=\dfrac{-12}{3}=-4$

Calculate the value of f(x) for x = -4:

$f(4)=3(-4)+12=-12+12=0$