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

+4
90
3
+108

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

Any help would be appreciated.

Aug 18, 2021

#1
+114818
+3

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}

Aug 18, 2021
#2
+108
-4

Thanks!

It looks like I made a miscalculation before in one of my steps

OrangeJuicy  Aug 18, 2021
#3
+114818
+1

You are welcome :)

Melody  Aug 18, 2021