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 \leq 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) = \frac{7 x ^{2} + 9 x - 10}{x ^{3}}

f (x) = 12 x - 2 ln (x), x > 0

(x^{2} - 9)^{6}

\frac{( 2 x - 1 )}{x ^{2} - x - 2}

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

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