What is the inverse of f(x)=9x^2-12,x>= 0 ?

The inverse of f(x)=9x^2-12,x>= 0 is (sqrt(x+12))/3 ,-(sqrt(x+12))/3

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inverse of f(x)=9x^2-12,x>= 0

inverse

f (x) = 9 x^{2} - 12, x \geq 0

\frac{x + 12}{3}, - \frac{x + 12}{3}

Solution steps

y = 9 x^{2} - 12

Replace

x

with

y

x = 9 y^{2} - 12

Solve for

y, x = 9 y^{2} - 12

\frac{x + 12}{3}, - \frac{x + 12}{3}

f (x) = \frac{1}{x + 5}

\frac{3 x + 10}{x + 3}

f (x) = \frac{3 x + 2}{5 x - 3}

\frac{2 + 4 x}{- 4 - 5 x}

y = \frac{( 3 x - 7 )}{( 9 x + 1 )}

What is the inverse of f(x)=9x^2-12,x>= 0 ?
The inverse of f(x)=9x^2-12,x>= 0 is (sqrt(x+12))/3 ,-(sqrt(x+12))/3