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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

x^{3} + 3 x + 8

f (x) = \frac{1}{x + 11}

sec (x)

f (x) = x^{3} - 7 x + 5, 0 \leq x \leq 3

y = - 1.75 (0) + 19

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