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,


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




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 $x$?

  • $\begingroup$ 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 $(-\infty, -\frac{1}{4}]$. $\endgroup$ – N. F. Taussig Jun 23 '15 at 13:05
  • $\begingroup$ I think the question asked me to solve the value of $x$ such that $g(x)=g^{-1}(x) $\endgroup$ – Mathxx Jun 23 '15 at 13:24
  • $\begingroup$ @N.F.Taussig how? $\endgroup$ – Mathxx Jun 23 '15 at 13:52
  • $\begingroup$ Setting $g(x) = g^{-1}(x)$ gives $$3 - 2x - 4x^2 = \frac{-\sqrt{13 - 4x} - 1}{4}$$ Multiplying both sides of the equation by $4$, then adding $1$ to each side of the resulting equation yields $$13 - 8x - 16x^2 = -\sqrt{13 - 4x}$$ Squaring each side of the resulting equation yields $$169 + 64x^2 + 256x^4 - 208x - 416x^2 + 256x^3 = 13 - 4x$$ Adding $4x - 13$ to each side of the equation and simplifying yields $$256x^4 + 256x^3 - 352x^2 - 204x + 156 = 0$$ You can try checking for rational roots, but you may have to use numerical methods (or a graphing calculator) to solve for $x$. $\endgroup$ – N. F. Taussig Jun 23 '15 at 14:31
  • $\begingroup$ I got $64x^4+64x^3-88x^2-51x+39=0$ can I factorise it ? $\endgroup$ – Mathxx Jun 23 '15 at 14:41

The secret to solving where $g(x)=g^{-1}(x)$ lies in the realization that you do not need to set the two functions equal to one another and solve a complicated quartic equation.

Recall that, geometrically speaking, the inverse $g^{-1}$ of $g$ is the reflection of $g$ over the line $y=x$. This being the case, the only way that $g(x)=g^{-1}(x)$ is if $g(x)=x$ and $g^{-1}(x)=x$. If you are having trouble understanding what I'm saying here draw the graph of an invertible function $f$, then on the same figure draw the graph of the line $y=x$, and see what happens to points reflected over that line.

Setting $g(x)=x$ gives the equation $4x^2+3x-3=0$. The solution, taking the domain into account, is $$x=\frac{-3-\sqrt{57}}{8}.$$

You should note that setting $g^{-1}(x)=x$ gives precisely the same quadratic equation as above.


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