What is the inverse of f(x)=9x^2-3 ?

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

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inverse of f(x)=9x^2-3

inverse

f (x) = 9 x^{2} - 3

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

Solution steps

y = 9 x^{2} - 3

Replace

x

with

y

x = 9 y^{2} - 3

Solve for

y, x = 9 y^{2} - 3

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

f (x) = e^{2 x} + 5

f (x) = - 7 cos (2 x), 0 \leq x \leq \frac{π}{2}

lo g_{\frac{5}{4}} (5 x - 4 x^{2})

f (x) = \frac{- 10 x + 1}{9 x + 6}

y = 5 (0.5)^{- x}

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