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

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

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

inverse

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

- \frac{x}{2}, - - \frac{x}{2}

Solution steps

y = - 2 x^{2}

Replace

x

with

y

x = - 2 y^{2}

Solve for

y, x = - 2 y^{2}

- \frac{x}{2}, - - \frac{x}{2}

\frac{x - 3}{5 x - 15}

f (x) = x + 4

(- 9, 3), (7, - 8)

y = \frac{1}{x + 4}

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

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