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

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

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

inverse

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

\frac{x ^{2} + 1}{2}, - \frac{x ^{2} + 1}{2}

Solution steps

y = 2 x^{2} - 1

Replace

x

with

y

x = 2 y^{2} - 1

Solve for

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

\frac{x ^{2} + 1}{2}, - \frac{x ^{2} + 1}{2}

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

f (x) = 99 \cdot (\frac{1}{2})^{x}

ln (\frac{( x - 3 )}{x - 1})

\frac{sin ^{2} ( 4 x )}{[ 5 x \cdot sin ( 3 x ) ]}

y = x^{2} - 12

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