What is the inverse of f(x)=(18x+1)^2 ?

The inverse of f(x)=(18x+1)^2 is (sqrt(x)-1)/(18),(-sqrt(x)-1)/(18)

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inverse of f(x)=(18x+1)^2

inverse

f (x) = (18 x + 1)^{2}

\frac{x - 1}{18}, \frac{- x - 1}{18}

Solution steps

y = (18 x + 1)^{2}

Replace

x

with

y

x = (18 y + 1)^{2}

Solve for

y, x = (18 y + 1)^{2}

\frac{x - 1}{18}, \frac{- x - 1}{18}

d o main x^{3} - 2

a sy m pt o t es 3 - 2 x - x^{2}

a sy m pt o t es f (x) = \frac{9 x ^{2} + 36 x + 41}{3 x + 5}

in v erse f (x) = (x - 15)^{2}, x \leq 15

d o main f (x) = 20 x^{3}

What is the inverse of f(x)=(18x+1)^2 ?
The inverse of f(x)=(18x+1)^2 is (sqrt(x)-1)/(18),(-sqrt(x)-1)/(18)