What is the inverse of f(x)=4x^2-1,x<= 0 ?

The inverse of f(x)=4x^2-1,x<= 0 is (sqrt(x+1))/2 ,-(sqrt(x+1))/2

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inverse of f(x)=4x^2-1,x<= 0

inverse

f (x) = 4 x^{2} - 1, x \leq 0

\frac{x + 1}{2}, - \frac{x + 1}{2}

Solution steps

y = 4 x^{2} - 1

Replace

x

with

y

x = 4 y^{2} - 1

Solve for

y, x = 4 y^{2} - 1

\frac{x + 1}{2}, - \frac{x + 1}{2}

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

f (x) = \frac{2}{x}

f (x) = \frac{- 10 x + 20}{x ^{2} - 3 x - 10}

f (x) = x^{3} - 3 x + 3

f (x) = \frac{5 x - 10}{3 x - 15}

What is the inverse of f(x)=4x^2-1,x<= 0 ?
The inverse of f(x)=4x^2-1,x<= 0 is (sqrt(x+1))/2 ,-(sqrt(x+1))/2