What is the inverse of f(x)=4x^2+16x-3 ?

The inverse of f(x)=4x^2+16x-3 is (-4+sqrt(x+19))/2 ,-(sqrt(x+19)+4)/2

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inverse of f(x)=4x^2+16x-3

inverse

f (x) = 4 x^{2} + 16 x - 3

\frac{- 4 + x + 19}{2}, - \frac{x + 19 + 4}{2}

Solution steps

y = 4 x^{2} + 16 x - 3

Replace

x

with

y

x = 4 y^{2} + 16 y - 3

Solve for

y, x = 4 y^{2} + 16 y - 3

\frac{- 4 + x + 19}{2}, - \frac{x + 19 + 4}{2}

(\frac{1}{4}, - \frac{1}{2}), (\frac{3}{4}, 2)

\frac{x ^{2} - 7 x + 10}{x + 2}

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

y = \frac{x ^{2} - x}{x ^{2} - 5 x + 4}

y = f (x) = ln (2 x + 1) - 2 x - 1

What is the inverse of f(x)=4x^2+16x-3 ?
The inverse of f(x)=4x^2+16x-3 is (-4+sqrt(x+19))/2 ,-(sqrt(x+19)+4)/2