Science:Math Exam Resources/Courses/MATH100/December 2015/Question 10 (g)

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MATH100 December 2015

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Question 10 (g)
Let $f(x)=(x+2)(x+1)e^{-x}$ . (g) Find the intervals on which $f(x)$ is increasing. Find the intervals on which $f(x)$ is decreasing.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A function is increasing when $f'(x)>0$ and decreasing when $f'(x)<0$ .

Hint 2
In the expression of $f'(x)$ which part really determines its sign.

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Solution

We need to determine the sign of $f'(x)=-e^{-x}\left(x^{2}+x-1\right)$ .

From part (e) we have $f'(x)=-e^{-x}\left(x^{2}+x-1\right)$ , where $e^{-x}$ is always positive so it doesn't have any role in the sign. We only need to determine the sign of $-x^{2}-x+1$ .

From part (f) $-x^{2}-x+1$ is $0\ \quad$ at $x={\dfrac {-1+{\sqrt {5}}}{2}}\approx 0.35,\ {\text{and}}\ \ x={\dfrac {-1-{\sqrt {5}}}{2}}\approx -0.13$ ,

so we make a sign chart for $f'(x)$ to find where it is positive and negative and therefore where the function $f(x)$ is increasing and decreasing.

	$\displaystyle (-\infty ,{\dfrac {-1-{\sqrt {5}}}{2}})$	$\displaystyle ({\dfrac {-1-{\sqrt {5}}}{2}},{\dfrac {-1+{\sqrt {5}}}{2}})$	$\displaystyle ({\dfrac {-1+{\sqrt {5}}}{2}},\infty )$
$-x^{2}-x+1$	$\displaystyle -$	$\displaystyle +$	$\displaystyle -$
$f'(x)=e^{-x}\left(-x^{2}-x+1\right)$	$\displaystyle -$	$\displaystyle +$	$\displaystyle -$
$f(x)$	decreasing	increasing	decreasing