What is the inverse of f(x)=x^2-4,-3<x<=-1 ?

The inverse of f(x)=x^2-4,-3<x<=-1 is sqrt(x+4),-sqrt(x+4)

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inverse of f(x)=x^2-4,-3<x<=-1

inverse

f (x) = x^{2} - 4, - 3 < x \leq - 1

x + 4, - x + 4

Solution steps

y = x^{2} - 4

Replace

x

with

y

x = y^{2} - 4

Solve for

y, x = y^{2} - 4

x + 4, - x + 4

3.67

sec^{2} (\frac{x}{1 - x ^{2}})

e^{x + 9} - 7

(x - 19)^{2}

f (x) = 1 0^{x} + \frac{2}{3}

What is the inverse of f(x)=x^2-4,-3<x<=-1 ?
The inverse of f(x)=x^2-4,-3<x<=-1 is sqrt(x+4),-sqrt(x+4)