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

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

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inverse of f(x)=-2x^2+8

inverse

f (x) = - 2 x^{2} + 8

- \frac{x - 8}{2}, - - \frac{x - 8}{2}

Solution steps

y = - 2 x^{2} + 8

Replace

x

with

y

x = - 2 y^{2} + 8

Solve for

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

- \frac{x - 8}{2}, - - \frac{x - 8}{2}

9 x - 8

f (x) = arcsin (\frac{( 2 x - 1 )}{3})

f (x) = \frac{( x + 5 )}{x}

f (x) = 3 sin (x)

y = x^{9}

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