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# inverses

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find the inverse of f(x)=3/5x^3-9. Then verify that f(x) and f^-1(x) are truly inverses

May 25, 2022

#1
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Hey can you put the parenthesis in the right spots?

May 25, 2022
#2
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hi, they are in the right spots. is there something from your question i misunderstood?

xxJenny1213xx  May 25, 2022
#3
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wait is it (3)/(5x^3)-9?

hipie  May 25, 2022
#4
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3/5 (x^3-9) i believe

xxJenny1213xx  May 25, 2022
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Your notation is fine Jenny,     (although brackets do cut down any possibility of confusion)

f(x)=3/5x^3-9

$$f(x)=\frac{3}{5}x^3-9\\ let \;\;y=f(x)\\ y=\frac{3}{5}x^3-9\\ \text{make x the subject}\\ 5y=3x^3-45\\ \frac{5y+45}{3}=x^3\\ x=\sqrt[3]{\frac{5y+45}{3}}\\ so\\ f^{-1}(x)=\sqrt[3]{\frac{5x+45}{3}}\\$$

I don't know what they want for the verification ....

Here is graph validation I guess

A function and its inverse are reflections of one another over the line y=x

Latex

f(x)=\frac{3}{5}x^3-9\\
let \;\;y=f(x)\\
y=\frac{3}{5}x^3-9\\
\text{make x the subject}\\
5y=3x^3-45\\
\frac{5y+45}{3}=x^3\\
x=\sqrt[3]{\frac{5y+45}{3}}\\
so\\
f^{-1}(x)=\sqrt[3]{\frac{5x+45}{3}}\\

May 25, 2022
edited by Melody  May 25, 2022
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thank you!

xxJenny1213xx  May 25, 2022
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You are very welcome.

Melody  May 25, 2022