What is the inverse of f(x)=9x^2,x>= 0 ?

The inverse of f(x)=9x^2,x>= 0 is (sqrt(x))/3 ,-(sqrt(x))/3

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inverse of f(x)=9x^2,x>= 0

inverse

f (x) = 9 x^{2}, x \geq 0

\frac{x}{3}, - \frac{x}{3}

Solution steps

y = 9 x^{2}

Replace

x

with

y

x = 9 y^{2}

Solve for

y, x = 9 y^{2}

\frac{x}{3}, - \frac{x}{3}

f (x) = x^{7} + 3

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

f (x) = - 4 (x - 2)^{2} - 3

f (x) = 2 x^{3} + 1

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

What is the inverse of f(x)=9x^2,x>= 0 ?
The inverse of f(x)=9x^2,x>= 0 is (sqrt(x))/3 ,-(sqrt(x))/3