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

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

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

inverse

f (x) = \frac{8 x + 1}{16 x ^{2} - 4}

\frac{1 + - x ( - 4 x - 1 ) + 1}{4 x}, \frac{1 - - x ( - 4 x - 1 ) + 1}{4 x}

Solution steps

y = \frac{8 x + 1}{16 x ^{2} - 4}

Replace

x

with

y

x = \frac{8 y + 1}{16 y ^{2} - 4}

Solve for

y, x = \frac{8 y + 1}{16 y ^{2} - 4}

\frac{1 + - x ( - 4 x - 1 ) + 1}{4 x}, \frac{1 - - x ( - 4 x - 1 ) + 1}{4 x}

y = sin (6 x)

f (x) = 3 x + 11

f (x) = 3 x + 18

cos (- \frac{5}{6 7})

f (x) = - (ln (1 - x))

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