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What is the inverse of [tex]$f(x)=3 x^3-4$[/tex]?

A. [tex]f^{-1}(x)=\frac{x+1}{3}[/tex]

B. [tex]f^{-1}(x)=\sqrt[2]{\frac{x+4}{3}}[/tex]

C. [tex]f^{-1}(x)=3 x+4[/tex]

D. [tex]f^{-1}(x)=\sqrt{\frac{x+4}{3}}[/tex]


Sagot :

To find the inverse of the function [tex]\( f(x) = 3x^3 - 4 \)[/tex], we need to follow a systematic approach:

1. Start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 3x^3 - 4 \][/tex]

2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:

- Step 1: Add 4 to both sides of the equation:
[tex]\[ y + 4 = 3x^3 \][/tex]

- Step 2: Divide both sides by 3:
[tex]\[ \frac{y + 4}{3} = x^3 \][/tex]

- Step 3: Take the cube root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \left( \frac{y + 4}{3} \right)^{\frac{1}{3}} \][/tex]

3. Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to express the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \left( \frac{x + 4}{3} \right)^{\frac{1}{3}} \][/tex]

Thus, the inverse function is [tex]\( f^{-1}(x) = \left( \frac{x + 4}{3} \right)^{\frac{1}{3}} \)[/tex].

Comparing this with the given options:
- A) [tex]\( f^{-1}(x) = \frac{x+1}{3} \)[/tex]
- B) [tex]\( f^{-1}(x) = \sqrt[2]{\frac{x+4}{3}} \)[/tex]
- C) [tex]\( f^{-1}(x) = 3x + 4 \)[/tex]
- D) [tex]\( f^{-1}(x) = \sqrt{\frac{x+4}{3}} \)[/tex]

We see that none of the options exactly match the derived inverse. The correct expression derived for the inverse function does not perfectly align with any of the provided multiple choices.