# Inverse function $g^{-1}$

The function $g$ is defined by $$g(x)= 3-2x-4x^2, x\in \mathbb{R},x\leq -\frac{1}{4}$$

Find the inverse function $g^{-1}$. Calculate the value of $x$ for which $g(x)=g^{-1}x$.

My attempt,

$g(x)=3-2x-4x^2$

Let $g^{-1}(x)=1$

$x=g(a)$

$=3-2a-4a^2$

$0=3-2a-4a^2-x$

Solving for $a$, I got $\frac{\pm \sqrt{13-4x}-1}{4}$

Since $x\leq -\frac{1}{4}$, so $f^{-1}(x)=\frac{-\sqrt{13-4x}-1}{4}$

Am I correct for my inverse of $g(x)$? How to proceed to find the value of $x$?

• You correctly solved for $g^{-1}$. However, the rest of your question is unclear. Do you wish to solve for the value of $x$ such that $g(x) = g^{-1}(x)$? Why did you set $g^{-1}(x) = 1$? Do you wish to solve for $x$ such that $g^{-1}(x) = 1$? No such $x$ exists since the range of $g^{-1}(x)$ is $(-\infty, -\frac{1}{4}]$. – N. F. Taussig Jun 23 '15 at 13:05