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

The inverse of f(x)=(8x)/(x^2+1) is (4+sqrt(-x^2+16))/x ,(4-sqrt(-x^2+16))/x

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

inverse

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

\frac{4 + - x ^{2} + 16}{x}, \frac{4 - - x ^{2} + 16}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{4 + - x ^{2} + 16}{x}, \frac{4 - - x ^{2} + 16}{x}

f (x) = - 3 x^{2} + 36 x

f

m = - 4, (6, 5)

tan (2 x - 5)

f (x) = - 25 x^{2} - 10 x - 1

What is the inverse of f(x)=(8x)/(x^2+1) ?
The inverse of f(x)=(8x)/(x^2+1) is (4+sqrt(-x^2+16))/x ,(4-sqrt(-x^2+16))/x