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}

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