What is the inverse of f(x)=2x^2-1,x>= 0 ?

The inverse of f(x)=2x^2-1,x>= 0 is sqrt((x+1)/2),-sqrt((x+1)/2)

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inverse of f(x)=2x^2-1,x>= 0

inverse

f (x) = 2 x^{2} - 1, x \geq 0

\frac{x + 1}{2}, - \frac{x + 1}{2}

Solution steps

y = 2 x^{2} - 1

Replace

x

with

y

x = 2 y^{2} - 1

Solve for

y, x = 2 y^{2} - 1

\frac{x + 1}{2}, - \frac{x + 1}{2}

f (x) = 2^{7 x - 6}

f (x) = 5 x - 15

x^{7} + 3

f (x) = 2 - \frac{12}{x + 5}

f (x) = 3.125 \cdot x

What is the inverse of f(x)=2x^2-1,x>= 0 ?
The inverse of f(x)=2x^2-1,x>= 0 is sqrt((x+1)/2),-sqrt((x+1)/2)