Question

Given that f(x)= x^3/4 +6 a) Find f(4) b) Find f^-1(x) c) Find f^-1(8)

Answer 1

a)

$f(4)=\dfrac{4^3}{4}+6=4^2+6=16+6=22$

b)

$\begin{aligned}\\&y=\dfrac{x^3}{4}+6\\&4y=x^3+24\\&x^3=4y-24\\&x=\sqrt[3]{4y-24}\\&f^{-1}(x)=\sqrt[3]{4x-24}\end$

c)

$f^{-1}(8)=\sqrt[3]{4\cdot8-24}=\sqrt[3]{8}=2$

Answer 2

Answer:

$\displaystyle \large{f(4)=22}\\\\\displaystyle \large{f^{-1}(x)=\sqrt[3]{4x-24}}\\\\\displaystyle \large{f^{-1}(8) = 2}$

Step-by-step explanation:

In this problem, we are given the linear function:

$\displaystyle \large{f(x)=\dfrac{x^3}{4}+6}$

( a ) Find f(4)

Simply substitute x = 4 in the function f(x).

$\displaystyle \large{f(4)=\dfrac{4^3}{4}+6}\\\\\displaystyle \large{f(4)=4^2+6}\\\\\displaystyle \large{f(4)=16+6}\\\\\displaystyle \large{f(4)=22}$

( b ) Find the inverse

To find $\displaystyle \large{f^{-1}(x)}$, solve for x-term then swap x-term and y-term.

$\displaystyle \large{4f(x)=x^3+24}\\\\\displaystyle \large{4f(x)-24=x^3}\\\\\displaystyle \large{\sqrt[3]{4f(x)-24}=\sqrt[3]{x^3}}\\\\\displaystyle \large{x=\sqrt[3]{4f(x)-24}}$

Swap f(x) and x.

$\displaystyle \large{f^{-1}(x)=\sqrt[3]{4x-24}}$

( c ) Find inverse f(8)

Substitute x = 8 in inverse function.

$\displaystyle \large{f^{-1}(8) = \sqrt[3]{4(8)-24}}\\\\\displaystyle \large{f^{-1}(8) = \sqrt[3]{32-24}}\\\\\displaystyle \large{f^{-1}(8) = \sqrt[3]{8}}\\\\\displaystyle \large{f^{-1}(8) = 2}$

Please let me know if you have any doubts!