[tex]B)f^{-1}(x)=\sqrt[3]{\frac{x+4}{3}}[/tex]
Explanation
[tex]f(x)=3x^3-4[/tex]
Step 1
swap the variables x and y
[tex]\begin{gathered} f(x)=3x^3-4 \\ y=3x^3-4 \\ y=3x^3-4\rightarrow x=3y^3-4 \\ x=3y^3-4 \end{gathered}[/tex]
Step 2
now, isolate y
[tex]\begin{gathered} x=3y^3-4 \\ \text{add 4 in both sides} \\ x+4=3y^3-4+4 \\ x+4=3y^3 \\ \text{divide both sides by 3} \\ \frac{x+4}{3}=\frac{3y^3}{3} \\ \frac{x+4}{3}=y^3 \\ \text{get the cubic root in both sides} \\ \sqrt[3]{\frac{x+4}{3}}=\sqrt[3]{y^3} \\ \sqrt[3]{\frac{x+4}{3}}=y\rightarrow inverse\text{ function} \end{gathered}[/tex]
therefore, the answer is
[tex]B)f^{-1}(x)=\sqrt[3]{\frac{x+4}{3}}[/tex]
I hope this helps you
