What is the inverse of f(x)=x^2+1,x>0 ?

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

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inverse of f(x)=x^2+1,x>0

inverse

f (x) = x^{2} + 1, x > 0

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

51900000

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

f (x) = 2 + ln (3 x + 4)

f (x) = 2^{x + 3} - 1

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

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