To find number of real solutions of equation $\left(\frac {9}{10}\right)^x =-3+x-x^2 $

Question

To find real solutions of

$\left(\frac{9}{10}\right)^x = -3 + x -x^2$

I differentiate it to get

$\left(\frac {9}{10}\right)^x log(\frac{9}{10})-1 + 2x=0 $

As x goes to $+\infty$ this goes to $+\infty$ and as $x$ goes to $-\infty$ it goes to $-\infty$. So real root should be 1. But i am not sure whether it has more than 1.

Thanks

when $x$ goes to negative infinity,how do you say your expression on LHS goes to negative infinity — Ekaveera Gouribhatla, Jan 01 '16 at 07:23

score 5 · Accepted Answer · answered Jan 01 '16 at 07:16

5

The function $-3+x-x^2$ is always negative, so there are no solutions.

answered Jan 01 '16 at 07:16

pre-kidney

30,223

But where is my mistake – Sophie Clad Jan 01 '16 at 07:18
You worked with the wrong function. Your reasoning shows that $f'(x)$ has a root. But that doesn't matter if you want to find roots of $f(x)$. – pre-kidney Jan 01 '16 at 07:24
1

@SophieClad The limit as $x\rightarrow-\infty$ is not $-\infty$ because $(9/10)^x$ actually grows as $x\rightarrow-\infty$. – Gregory Grant Jan 01 '16 at 07:25
Right, @GregoryGrant is correct that you also had a mistake in your analysis of $f'(x)$. So there are several layers of issues as it turns out. – pre-kidney Jan 01 '16 at 07:28

Claude Leibovici · Answer 2 · 2016-01-01T09:00:37.047

Beside pre-kidney's answer which is clear, considering the function $$f(x)=\left(\frac{9}{10}\right)^x +3 -x+x^2$$ you properly obtained for the derivative $$f'(x)=\left(\frac {9}{10}\right)^x \log(\frac{9}{10})-1 + 2x$$ I shall add that $$f''(x)=\left(\frac{9}{10}\right)^x \log ^2\left(\frac{10}{9}\right)+2 $$ which is always positive.

The derivative cancels close to $x=0.450$ but, for this value, $f(x)=3.706$ and $f''(x)=2.010$. SO, the function just goes through this minimum point.

Edit

You probably noticed that $f(x)$ looks more or less like a parabola. Just for fun, let us build the Taylor series around $x=0$. We obtain $$f(x)=4- \left(1+\log \left(\frac{10}{9}\right)\right)x+ \left(1+\frac{1}{2} \log ^2\left(\frac{10}{9}\right)\right)x^2+O\left(x^3\right)$$ Plot the function and this approximation for $-10\leq x \leq 10$; you will be amazed to see how close they are.

Happy New Year !

Happy New Year to you as well – Sophie Clad Jan 01 '16 at 08:57 — Sophie Clad, Jan 01 '16 at 08:57

To find number of real solutions of equation $\left(\frac {9}{10}\right)^x =-3+x-x^2 $

2 Answers2

Happy New Year !