What is the inverse of f(x)=-4x^2-5,x<0 ?

The inverse of f(x)=-4x^2-5,x<0 is sqrt(-(x+5)/4),-sqrt(-(x+5)/4)

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inverse of f(x)=-4x^2-5,x<0

inverse

f (x) = - 4 x^{2} - 5, x < 0

- \frac{x + 5}{4}, - - \frac{x + 5}{4}

Solution steps

y = - 4 x^{2} - 5

Replace

x

with

y

x = - 4 y^{2} - 5

Solve for

y, x = - 4 y^{2} - 5

- \frac{x + 5}{4}, - - \frac{x + 5}{4}

2 x - 10

4 x - 2

f (x) = \frac{3 ^{x}}{3 ^{x} + 1}

f (x) = arccos (- 1)

What is the inverse of f(x)=-4x^2-5,x<0 ?
The inverse of f(x)=-4x^2-5,x<0 is sqrt(-(x+5)/4),-sqrt(-(x+5)/4)