What is the inverse of f(x)=\sqrt[3]{2x^2-16} ?

The inverse of f(x)=\sqrt[3]{2x^2-16} is sqrt((x^3+16)/2),-sqrt((x^3+16)/2)

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inverse of f(x)=\sqrt[3]{2x^2-16}

inverse

\frac{x ^{3} + 16}{2}, - \frac{x ^{3} + 16}{2}

Solution steps

Replace

x

with

y

Solve for

\frac{x ^{3} + 16}{2}, - \frac{x ^{3} + 16}{2}

f (x) = 32 + y \frac{9}{5}

y = \frac{2 - x}{2 x - 1}

f (x) = \frac{- 6 - 7 x}{2}

f (x) = 2 x - 38

f (x) = 9 x^{2} + 8 x + 5

What is the inverse of f(x)=\sqrt[3]{2x^2-16} ?
The inverse of f(x)=\sqrt[3]{2x^2-16} is sqrt((x^3+16)/2),-sqrt((x^3+16)/2)