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

The inverse of f(x)= 1/(1+8x^2) is (sqrt(2)sqrt(x)sqrt(1-x))/(4x),-(sqrt(2)sqrt(x)sqrt(1-x))/(4x)

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

inverse

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

\frac{2 x 1 - x}{4 x}, - \frac{2 x 1 - x}{4 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{2 x 1 - x}{4 x}, - \frac{2 x 1 - x}{4 x}

arccsc (e^{x}) - 2

f (x) = 5^{2} + 600 x - 70

g (x) = 2 + 2 - x

f (x) = \frac{( 7 x - 2 )}{x + 9}

\frac{1}{( s + 2 ) ( s + 5 )}

What is the inverse of f(x)= 1/(1+8x^2) ?
The inverse of f(x)= 1/(1+8x^2) is (sqrt(2)sqrt(x)sqrt(1-x))/(4x),-(sqrt(2)sqrt(x)sqrt(1-x))/(4x)