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

Answer 1

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

Answer 2

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 !