What is the inverse of f(x)=3x^2+4x+1 ?

The inverse of f(x)=3x^2+4x+1 is (-2+sqrt(3x+1))/3 ,-(sqrt(3x+1)+2)/3

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inverse of f(x)=3x^2+4x+1

inverse

f (x) = 3 x^{2} + 4 x + 1

\frac{- 2 + 3 x + 1}{3}, - \frac{3 x + 1 + 2}{3}

Solution steps

y = 3 x^{2} + 4 x + 1

Replace

x

with

y

x = 3 y^{2} + 4 y + 1

Solve for

y, x = 3 y^{2} + 4 y + 1

\frac{- 2 + 3 x + 1}{3}, - \frac{3 x + 1 + 2}{3}

f (x) = x^{2} + 10 x + 16

f (x) = \frac{( 2 x - 3 )}{x}

f (x) = (7 - 14 x)^{\frac{1}{2}}

g (x) = 2 x^{2} - 8

f (x) = \frac{( x + 5 )}{2}

What is the inverse of f(x)=3x^2+4x+1 ?
The inverse of f(x)=3x^2+4x+1 is (-2+sqrt(3x+1))/3 ,-(sqrt(3x+1)+2)/3