What is the inverse of f(x)=sqrt(1+x^2) ?

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

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inverse of f(x)=sqrt(1+x^2)

inverse

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

x^{2} - 1, - x^{2} - 1

Solution steps

y = 1 + x^{2}

Replace

x

with

y

x = 1 + y^{2}

Solve for

y, x = 1 + y^{2}

x^{2} - 1, - x^{2} - 1

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

f (x) = \frac{2 - x ^{2}}{7}

f (x) = \frac{1}{x ^{2} - 9}

f (x) = 2 (x - 3)^{2} - 8

f (x) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

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