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

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

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

inverse

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

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

Solution steps

y = \frac{1}{2 x ^{2} + 2}

Replace

x

with

y

x = \frac{1}{2 y ^{2} + 2}

Solve for

y, x = \frac{1}{2 y ^{2} + 2}

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

f (x) = (5, - 2)

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

4 x^{4}

2 + x - 1

f (x) = 10 (x + 8)^{7} - 10

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