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I'm trying to find $f^{-1}(x)$ for $f(x) = -5x^2 + 10x + 2$. The correct answer is $f^{-1}(x) = 1 \pm \sqrt{1.4-0.2x}$, but I can't derive that.

What I've come up with is: $f^{-1}(x) = 1\pm\frac{\sqrt{140-20x}}{-10}$ which is close, but not exactly what should be the correct answer.

I got that by:

\begin{align*}f(x) &= -5x^2+10x+2\\ &\\ y &= -5x^2+10x+2\\ &\\ x &= -5y^2+10y+2\\ &\\ 0 &= -5y^2+10y+2-x\\ \end{align*}

Now plugging this into the quadratic formula gives:

$$\frac{-10\pm{\sqrt{10^2-4\times(-5)\times(2-x)}}}{2\times(-5)}$$

$$\frac{-10\pm{\sqrt{100+20\times(2-x)}}}{-10}$$

$$\frac{-10}{-10}\pm{\frac{\sqrt{140-20x}}{-10}}$$

$$1\pm{\frac{\sqrt{140-20x}}{-10}}$$

Now my question is how does one get from $\dfrac{\sqrt{140-20x}}{-10}$ to $\sqrt{1.4-0.2x}$? I could easily understand how to get there from $\dfrac{\sqrt{140-20x}}{10}$, because:

$$\frac{\sqrt{140-20x}}{10} = \frac{\sqrt{140-20x}}{\sqrt{100}} = \sqrt{\frac{140-20x}{100}} = \sqrt{\frac{140}{100}-\frac{20x}{100}} = \sqrt{1.4-0.2x}$$

but with $-10$ as denominator I'm simply lost. How do I get from $-10$ to $\sqrt{100}$? Is it all right to just forget the negative sign? And if it's ok, is it because there is a plus-minus already in front of all this?

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4 Answers 4

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Note that $$1\pm\frac{\sqrt{140-20x}}{-10} = 1\pm \frac{\sqrt{1.4-0.2x}}{-1} = 1\mp\sqrt{1.4-0.2x}.$$ However, the choice between $\pm$ and $\mp$ here is arbitrary, as we are just writing two expressions in a compact form. That is, $1\mp\sqrt{1.4-0.2x}$ and $1\pm\sqrt{1.4-0.2x}$ both denote the two expressions $1+\sqrt{1.4-0.2x}$ and $1-\sqrt{1.4-0.2x}$.

In certain situations, it is important whether you use $\pm$ or $\mp$, but in this one it doesn't. An example where it does is the expression $a\pm b\mp c$. This is a compact way of writing the two expressions $a+b -c$ and $a-b+c$; that is, you choose the top operation throughout or the bottom operation throughout. For this reason, $a\pm b\mp c$ is different from $a\pm b\pm c$ and $a\mp b\mp c$.

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Let $$y=f(x)=-5x^2+10x+2\implies x=f^{-1}(y)$$

Now, $$5x^2-10x+y-2=0\implies x=?$$

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  • $\begingroup$ I'm sorry to ask for clarification, but I'm too stupid to understand why this works. With signs reversed, I can plug this into the quadratic formula and get the right answer, but I fail to understand why signs get reversed. Would you mind explaining a bit more? $\endgroup$ Jan 11, 2014 at 10:57
  • $\begingroup$ @2305843008139952128, do you mean why $$y=-5x^2+10x+20\implies 5x^2-10x+y-2=0 ?$$ then its simple sign change due to change of sides $\endgroup$ Jan 11, 2014 at 10:59
  • $\begingroup$ Got it, thank you very much. $\endgroup$ Jan 11, 2014 at 11:00
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Yes, a ± number divided by a - number is still a ± number.
ex.  ((±6)/(-2))=±3
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Let’s represent $f(x)=-5(x^2-2x) + 2 = -5(x^2-2x+1-1)+2=-5((x-1)^2-1)+2=-5(x-1)^2+7$

$y=-5(x-1)^2+7$

So, after we can change x and y

$x=-5(y-1)^2+7$

$\frac {7-x}{5}=(y-1)^2$

$ \pm \sqrt {\frac {7-x}{5}}=y-1$

$\pm \sqrt{1.4-0.2x}+1=y$

Thats all)

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