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Given: $f(x)=(x^2+a)e^{-x}$

( $0\le x\le 2$)

Prove that for $0

My attempt:

$f'(x)=e^{-x}(-x^2+2x-a)$.

$f'(x)=0$ , so $-x^2+2x-a=0$

$x=1\pm\sqrt{1-a}$, for $a<1$.

But how can i show for $a>0$?

2 Answers 2

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Hint

As you know:

$$f'(x)=e^{-x}(-x^2+2x-a)$$

$$f'(x)=0 \rightarrow x=1 \pm \sqrt{1-a}$$

If $0

Now you have to calculate

$$f''(x)=e^{-x}(x^2-4x+a+2)$$

$$f''(x)>0 \rightarrow x<2-\sqrt{2-a} \quad \text{or}\quad x>2+\sqrt{2-a} \quad (1)$$

$$f''(x)<0 \rightarrow 2-\sqrt{2-a}

Can you finish?

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    It tells nothing about the critical points lie in $(0,2)$. – 2017-01-16
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    @szw1710: really?! If $0$1\pm \sqrt{1-a}$? – 2017-01-16
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    Arnaldo: of course, it's trvial, but not explicitly written. :-) – 2017-01-16
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    @szw1710: that is why it is a HINT! – 2017-01-16
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    so then you put the $x$ we find and see where it fits? – 2017-01-16
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    @mila you have to check if the values of $x$ that you found lies in the interval $(1)$ or $(2)$. If it lies in $(1)$ then it is a minimum, if it lies in $(2)$ it is a maximum. – 2017-01-16
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    @mila: is it clear? – 2017-01-16
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    yes thank you! i could also do: $f''(1\pm \sqrt{1-a})$ and then to see when its positive or negative right? you did it backwards : first $f''(X)>0,f''(x)<0$ then checked. – 2017-01-17
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    @mila: yes, you can! – 2017-01-17
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The condition $a>0$ is needed to guarantee both critical points lie in $(0,2)$. The condition $a<1$ is needed for a square root to exist.

We have $1-\sqrt{1-a}<1<2$ and $1+\sqrt{1-a}>0$, which is trivial. Now the inequality $1-\sqrt{1-a}>0$ is equivalent to $\sqrt{1-a}<1\iff 1-a<1\iff a>0$. That $1+\sqrt{1-a}<2$ we solve in the same way.

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    but how do you show/prove that? – 2017-01-16
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    I put it in my answer above. – 2017-01-16
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    why it's $(0,2)$ and not as in the question: $[0,2]$? – 2017-01-16