What is the inverse of f(x)=-8x^2-3,x>= 0 ?

The inverse of f(x)=-8x^2-3,x>= 0 is sqrt(-(x+3)/8),-sqrt(-(x+3)/8)

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inverse of f(x)=-8x^2-3,x>= 0

inverse

f (x) = - 8 x^{2} - 3, x \geq 0

- \frac{x + 3}{8}, - - \frac{x + 3}{8}

Solution steps

y = - 8 x^{2} - 3

Replace

x

with

y

x = - 8 y^{2} - 3

Solve for

y, x = - 8 y^{2} - 3

- \frac{x + 3}{8}, - - \frac{x + 3}{8}

\frac{x ^{2} - x}{x - 2}

f (x) = 5 e^{x}

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

f (x) = 5 x^{2} + 13

f (x) = \frac{8 x - 1}{4}

What is the inverse of f(x)=-8x^2-3,x>= 0 ?
The inverse of f(x)=-8x^2-3,x>= 0 is sqrt(-(x+3)/8),-sqrt(-(x+3)/8)