What is the inverse of f(x)= 1/(sqrt(x^2+8)) ?

The inverse of f(x)= 1/(sqrt(x^2+8)) is (sqrt(1-8x^2))/x ,-(sqrt(1-8x^2))/x

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inverse of f(x)= 1/(sqrt(x^2+8))

inverse

f (x) = \frac{1}{x ^{2} + 8}

\frac{1 - 8 x ^{2}}{x}, - \frac{1 - 8 x ^{2}}{x}

Solution steps

y = \frac{1}{x ^{2} + 8}

Replace

x

with

y

x = \frac{1}{y ^{2} + 8}

Solve for

y, x = \frac{1}{y ^{2} + 8}

\frac{1 - 8 x ^{2}}{x}, - \frac{1 - 8 x ^{2}}{x}

What is the inverse of f(x)= 1/(sqrt(x^2+8)) ?
The inverse of f(x)= 1/(sqrt(x^2+8)) is (sqrt(1-8x^2))/x ,-(sqrt(1-8x^2))/x