What is the inverse of f(x)=9x^2+1 ?

The inverse of f(x)=9x^2+1 is (sqrt(x-1))/3 ,-(sqrt(x-1))/3

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inverse of f(x)=9x^2+1

inverse

f (x) = 9 x^{2} + 1

\frac{x - 1}{3}, - \frac{x - 1}{3}

Solution steps

y = 9 x^{2} + 1

Replace

x

with

y

x = 9 y^{2} + 1

Solve for

y, x = 9 y^{2} + 1

\frac{x - 1}{3}, - \frac{x - 1}{3}

f (x) = 9 x^{2} + 8

f (x) = 9 x^{2} + 7

f (x) = x^{6} + 16

f (x) = \frac{x - 2}{4 x + 9}

What is the inverse of f(x)=9x^2+1 ?
The inverse of f(x)=9x^2+1 is (sqrt(x-1))/3 ,-(sqrt(x-1))/3