What is the inverse of f(x)=-2x^2+6,x>= 0 ?

The inverse of f(x)=-2x^2+6,x>= 0 is sqrt(-(x-6)/2),-sqrt(-(x-6)/2)

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inverse of f(x)=-2x^2+6,x>= 0

inverse

f (x) = - 2 x^{2} + 6, x \geq 0

- \frac{x - 6}{2}, - - \frac{x - 6}{2}

Solution steps

y = - 2 x^{2} + 6

Replace

x

with

y

x = - 2 y^{2} + 6

Solve for

y, x = - 2 y^{2} + 6

- \frac{x - 6}{2}, - - \frac{x - 6}{2}

x - 1 + x + 1

2 (3 x + 1) + 4

f (x) = 1 + 3^{x}

f (x) = \frac{x + 4}{x + 12}

f (x) = \frac{x + 4}{x + 14}

What is the inverse of f(x)=-2x^2+6,x>= 0 ?
The inverse of f(x)=-2x^2+6,x>= 0 is sqrt(-(x-6)/2),-sqrt(-(x-6)/2)