I attempted to just use the regular way to find the inverse and plug in but my answer is wrong

Let's see

 

\(f(x)=\frac{x^5-1}3, Find f^{-1}(-31/96) \)

 

\(f(x)=\frac{x^5-1}{3}\\ y=\frac{x^5-1}{3}\\ 3y=x^5-1\\ 3y+1=x^5\\ x=\sqrt[5]{3y+1}\\~\\ f^{-1}(x)=\sqrt[5]{3x+1} \)

 

 

\(f^{-1}(-31/96)\\ =\sqrt[5]{3*\frac{-31}{96}+1}\\ =\sqrt[5]{\frac{-31}{32}+1}\\ =\sqrt[5]{\frac{1}{32}}\\ =0.5\)

 

 

 

LaTex:

f(x)=\frac{x^5-1}{3}\\
y=\frac{x^5-1}{3}\\
3y=x^5-1\\
3y+1=x^5\\
x=\sqrt[5]{3y+1}\\~\\
f^{-1}(x)=\sqrt[5]{3x+1}