3
$\begingroup$

For an equation $(ax^2+bx+c)e^{-x}$ I must find the three coefficients $a$, $b$ and $c$ by looking at this plot

enter image description here

I have the feeling I need to find three equations in order to form a coherent system but I only find these two:

$$0=0^2a+0b+c$$ $$16=4a+2b+c$$

I easily find out that $b=8-2a$ and $c=0$, but then what? I've been working on it much more than it surely deserves.

$\endgroup$

2 Answers 2

8
$\begingroup$

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^{-x}$, then $$f'(x)=(2ax+b-ax^2-bx-c)e^{-x}$$ and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

$\endgroup$
5
$\begingroup$

You also have $f(x) \geq 0$ for any $x \in \mathbb{R}$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^{-x}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .