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

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

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

inverse

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

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

Solution steps

y = (8 x + 1)^{2}

Replace

x

with

y

x = (8 y + 1)^{2}

Solve for

y, x = (8 y + 1)^{2}

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

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

\frac{s}{s ^{2} + 2 s + 10}

f (x) = 4 sin (x) - 3

f (x) = \frac{7}{8 x}

f (x) = \frac{( 3 x - 4 )}{8 x + 1}

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