What is the inverse of f(x)=2x^2-3,x>= 0 ?

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

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inverse of f(x)=2x^2-3,x>= 0

inverse

f (x) = 2 x^{2} - 3, x \geq 0

\frac{x + 3}{2}, - \frac{x + 3}{2}

Solution steps

y = 2 x^{2} - 3

Replace

x

with

y

x = 2 y^{2} - 3

Solve for

y, x = 2 y^{2} - 3

\frac{x + 3}{2}, - \frac{x + 3}{2}

y - 4 = 0

\frac{4 + 8 x + 4 x ^{2}}{x ^{2} - 2 x + 1}

\frac{1}{x ( 7 x ^{2} - 5 )}

6 x - 4 y = 40

f (x) = x - 5

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