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

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

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

inverse

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

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

y = - x^{2} + 1

f (x) = - x + 3

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

\frac{x ^{2}}{x ^{3} + 3}

f (x) = \frac{7}{14 - x}

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