What is the inverse of f(x)= 1/3 x^2-9 ?

The inverse of f(x)= 1/3 x^2-9 is sqrt(3x+27),-sqrt(3x+27)

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inverse of f(x)= 1/3 x^2-9

inverse

f (x) = \frac{1}{3} x^{2} - 9

3 x + 27, - 3 x + 27

Solution steps

y = \frac{1}{3} x^{2} - 9

Replace

x

with

y

x = \frac{1}{3} y^{2} - 9

Solve for

y, x = \frac{1}{3} y^{2} - 9

3 x + 27, - 3 x + 27

y = \frac{cos ( x )}{2 + sin ( x )}

f (x) = \frac{x}{2 x ^{2} - 5} - x

- 3 cos (5 x)

\frac{x ^{2} + 4 x + 7}{x + 3}

f (x) = 7 - x + 5

What is the inverse of f(x)= 1/3 x^2-9 ?
The inverse of f(x)= 1/3 x^2-9 is sqrt(3x+27),-sqrt(3x+27)