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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

lo g_{2} (x - 1) + 3

y = - lo g_{10} (1.074 x + 7) + 2

\frac{d}{d} (- x - 4)^{\frac{1}{4}}

y = \frac{4 x - 7}{2 x + 3}

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

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