What is the inverse of f(x)=-16x^2+40 ?

The inverse of f(x)=-16x^2+40 is sqrt(-(x-40)/(16)),-sqrt(-(x-40)/(16))

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inverse of f(x)=-16x^2+40

inverse

f (x) = - 16 x^{2} + 40

- \frac{x - 40}{16}, - - \frac{x - 40}{16}

Solution steps

y = - 16 x^{2} + 40

Replace

x

with

y

x = - 16 y^{2} + 40

Solve for

y, x = - 16 y^{2} + 40

- \frac{x - 40}{16}, - - \frac{x - 40}{16}

2 x^{2} - 5 x + 1

(- 1, 3), (4, - 2)

f (x) = \frac{- 9}{x ^{2} - 3}

y = \frac{1}{2} x + 8

f (x) = \frac{x + 7}{x ^{2} - 49}

What is the inverse of f(x)=-16x^2+40 ?
The inverse of f(x)=-16x^2+40 is sqrt(-(x-40)/(16)),-sqrt(-(x-40)/(16))