What is the inverse of h(x)=[(x)]^2+1 ?

The inverse of h(x)=[(x)]^2+1 is sqrt(x-1),-sqrt(x-1)

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inverse of h(x)=[(x)]^2+1

inverse

h (x) = [(x)]^{2} + 1

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

f (x) = \frac{- 5 x + 4}{3 x - 2}

ln (x^{2} - 3 x + 2)

y = 5

y = 7

f (x) = \frac{( 2 x - 3 )}{5}

What is the inverse of h(x)=[(x)]^2+1 ?
The inverse of h(x)=[(x)]^2+1 is sqrt(x-1),-sqrt(x-1)