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}

f (x) = 7 x^{3} - 12

f (x) = \frac{4}{5 x + 3}, x \neq = - \frac{3}{5}

\frac{1}{2} - x

f (x) = x^{2} + 15 x + 36

y = - 4.9 (t + 3)^{2} + 45.8

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)