What is the inverse of f(x)=16x^2,x>= 0 ?

The inverse of f(x)=16x^2,x>= 0 is (sqrt(x))/4 ,-(sqrt(x))/4

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inverse of f(x)=16x^2,x>= 0

inverse

f (x) = 16 x^{2}, x \geq 0

\frac{x}{4}, - \frac{x}{4}

Solution steps

y = 16 x^{2}

Replace

x

with

y

x = 16 y^{2}

Solve for

y, x = 16 y^{2}

\frac{x}{4}, - \frac{x}{4}

f (x) = \frac{x + 6}{5 - x}

f (x) = \frac{3}{x} - 7

f (x) = \frac{12}{7 - 6 x}

f (x) = \frac{1}{2} x + \frac{1}{4}

f (x) = \frac{( x + 4 )}{x + 5}

What is the inverse of f(x)=16x^2,x>= 0 ?
The inverse of f(x)=16x^2,x>= 0 is (sqrt(x))/4 ,-(sqrt(x))/4