$f(x)=e^x -10x^2$ doesn't vanish in more than three points. How can I prove that $f(x) = e^x - 10x^2$ doesn't vanish in more than three points.
I stuck here I just computed the derivative that is $f'(x) = e^x - 20x$ and then $x= \log(20)+\log(x)$ when $f'(x)=0$.
How can I prove this? Thanks for your help and time.
 A: $$e^x-10x^2=0$$ means 1. $$\sqrt{e}^x=\sqrt{10}x$$ and 2. 
$$\sqrt{e}^x=-\sqrt{10}x$$ 
Look at the function $y=x^{\frac{1}{x}}$ and compare with:
1.: $$1<(\sqrt{e}^x)^{(\sqrt{e}^{-x})}=\sqrt{e}^\sqrt{10}<e^{\frac{1}{e}}$$ => two Solutions
2.: $$0<(\sqrt{e}^x)^{(\sqrt{e}^{-x})}=\sqrt{e}^{-\sqrt{10}}<1$$ => one solution
Together: three solutions
A: We know that $$f(x) = e^x-10x^2 \\ f'(x) = e^x-20x \\ f''(x) = e^x-20$$Taking the second derivative and setting it equal to $0$ we get $$0=e^x-20 \\ \ln(20) = x$$ Therefore there is exactly one point where the second derivative is $0$. We also know that $$f''(x<\ln(20))<0\\f''(x>\ln(20))>0$$ What is the significance of this? This is the only point where the slope of the function can change from decreasing to increasing. Since $$f(\ln (20)) = 20 - 10*\ln(20)^2 < 0$$ and $$f'(\ln(20))=20-10*\ln(20) <0$$ There can only be $1$ zero to the right of $x=\ln(20)$. Since to the left of $x=\ln(20), f'(x)$ is decreasing, there can be at most $2$ zeroes to the left of $x=\ln(20)$. Therefore, the function has at most $3$ zeroes.
A: First try to localize the zeros for your function approximately. Look for changes of the sign of your function, use the intermediate value theorem to argue that there are zeros. Arrange the position of these zeros $x_1,x_2$ & $x_3$.
Then use the derivative to show that the function is strictly increasing/decreasing for the smalles zero and the largest zero, to conclude that there are no other zeros.
