What is the inverse of f(x)=-8x^2+4,x>= 0 ?

The inverse of f(x)=-8x^2+4,x>= 0 is sqrt(-(x-4)/8),-sqrt(-(x-4)/8)

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inverse of f(x)=-8x^2+4,x>= 0

inverse

f (x) = - 8 x^{2} + 4, x \geq 0

- \frac{x - 4}{8}, - - \frac{x - 4}{8}

Solution steps

y = - 8 x^{2} + 4

Replace

x

with

y

x = - 8 y^{2} + 4

Solve for

y, x = - 8 y^{2} + 4

- \frac{x - 4}{8}, - - \frac{x - 4}{8}

f (x) = \frac{1}{2} (x + 5) (x - 3) = 0

e^{x} + 5

f (x) = \frac{( x ^{2} - 3 )}{x + 2}

\frac{2}{x ^{2} + 2 x}

(- 4, - 1), (- 2, 5)

What is the inverse of f(x)=-8x^2+4,x>= 0 ?
The inverse of f(x)=-8x^2+4,x>= 0 is sqrt(-(x-4)/8),-sqrt(-(x-4)/8)