What is the inverse of f(x)=x^2-1,3<= x<5 ?

The inverse of f(x)=x^2-1,3<= x<5 is sqrt(x+1),-sqrt(x+1)

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inverse of f(x)=x^2-1,3<= x<5

inverse

f (x) = x^{2} - 1, 3 \leq x < 5

x + 1, - x + 1

Solution steps

y = x^{2} - 1

Replace

x

with

y

x = y^{2} - 1

Solve for

y, x = y^{2} - 1

x + 1, - x + 1

p (x) = x - 4

f (x) = 8 - y + 9

\frac{x - 11}{16}

\frac{x}{4} + 2

y^{\frac{3}{8}}

What is the inverse of f(x)=x^2-1,3<= x<5 ?
The inverse of f(x)=x^2-1,3<= x<5 is sqrt(x+1),-sqrt(x+1)