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How can you find the inverse of $f(x) = 2x^2+8x+13?$ This is what I've tried so far:

  1. $y = 2x^2+8x+13$
  2. $x = 2y^2+8y+13$
  3. $x-13 = 2y^2+8y$
  4. $x-13=y(y+8)$

This is where I got stuck. To be clear, I want to write $x$ in terms of $y.$ Credit to Jenna for fixing this.

Another way I tried to do it was the quadratic formula on the right side after step $2$ but I ended up with a discriminant of $-40.$

Edit: I meant inverse, not inverse function specifically. Sorry for asking the wrong thing. Should I delete and post another question or just keep this the way it is, edited?

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3 Answers 3

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I think you meant to write $x$ in terms of $y$? $$y=2x^2+8x+13$$$$y-13=2x^2+8x$$$$\frac{y-13}{2}=x^2+4x=(x+2)^2-4$$$$\frac{y-13}{2}+4=(x+2)^2$$$$\pm\sqrt{\frac{y-13}{2}+4}=x+2$$$$x=\pm\sqrt{\frac{y-13}{2}+4}-2$$

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This function does not have an inverse as it is not injective: check that $f(x-2) = f(-x-2)$ whatever $x$ is - for example, $f(1) = f(-5)$, etc.


We want to solve $y=2x^2+8x+13$ for $x$. Applying the quadratic formula for $2x^2+8x+(13-y)=0$ yields $$x = \frac{-8\pm \sqrt{64-4\cdot 2 \cdot (13-y)}}{4} = -2\pm \frac{\sqrt{16-26+2y}}{2} = -2\pm\frac{\sqrt{-10+2y}}{2}.$$The choice in $\pm$ depends on whether you pick $x \geq -2$ or $x \leq -2$ as your domain.

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  • $\begingroup$ Hi, would you mind checking out my edit? Thanks for your answer! $\endgroup$
    – Jack Pan
    Jun 19, 2016 at 20:56
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    $\begingroup$ You have to restrain your domain to $x \geq -2$ or $x \leq -2$ to get an inverse. One can find the inverse in such conditions by writing $2x^2+8x+(13-y)=0$ and applying the quadratic formula. $\endgroup$
    – Ivo Terek
    Jun 19, 2016 at 21:00
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    $\begingroup$ I elaborated a bit more. $\endgroup$
    – Ivo Terek
    Jun 19, 2016 at 21:05
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    $\begingroup$ Yeah, I had seen that just after I pressed the enter button, fixed already :P $\endgroup$
    – Ivo Terek
    Jun 19, 2016 at 21:09
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    $\begingroup$ You're absolutely right. Thanks. $\endgroup$
    – Ivo Terek
    Jun 19, 2016 at 21:12
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You can't find the inverse of a degree 2 polynomial defined on $R$ since it is not injective.

If you want an inverse on invervals, write $2x^2+8x+13=y$, $2x^2+8x+13-y=0$, $\Delta = 64-8(13-y)$ and $x={{-8+\sqrt{8(13-y)}}\over 4}$ or $x={{-8-\sqrt{8(13-y)}}\over 4}$

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  • $\begingroup$ Hi, would you mind checking out my edit? Thanks for your answer! $\endgroup$
    – Jack Pan
    Jun 19, 2016 at 20:57
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    $\begingroup$ If you want an inverse on invervals, write $2x^2+8x+13=y$, $2x^2+8x+13-y=0$, $\Delta = 64-8(13-y)$ and $x={{-8+\sqrt{8(13-y)}}\over 4}$ or $x={{-8-\sqrt{8(13-y)}}\over 4}$ $\endgroup$ Jun 19, 2016 at 21:01

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