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

MATH100 December 2015

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Question 10 (f)
Let $f(x)=(x+2)(x+1)e^{-x}$. (f) Find all critical points and singular points of $f(x)$.

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
Critical points of a function $f(x)$ are the points at which the derivative $f'(x)$ is either $0$ or NOT defined. In this question critical points are the $x$'s for which $f'(x)=0$, and singular points are those at which $f'(x)$ is NOT defined.

Hint 2
Given $f'(x)$ from part (e), is there any point at which $f'(x)$ is NOT defined?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From part (e), we have $f'(x)=-e^{-x}\left(x^{2}+x-1\right)$, which is the product of an exponential function and a polynomial, so it is defined everywhere, this means that there are NO singular points. The only critical points are then the points for which $f'(x)=-e^{-x}\left(x^{2}+x-1\right)=0$. $f'(x)=0\Rightarrow e^{-x}=0\quad {\text{or}}\quad x^{2}+x-1=0$ $e^{-x}$ is always positive so $e^{-x}=0\$ NEVER happens. We only need to solve $x^{2}+x-1=0$. We cannot factor, so we use the quadratic formula to find $x$. ${\begin{aligned}x^{2}+x-1=0&\Rightarrow x={\dfrac {-1\pm {\sqrt {1^{2}-4(1)(-1)}}}{2\times 1}}\\&\Rightarrow x={\dfrac {-1+{\sqrt {5}}}{2}},\ x={\dfrac {-1-{\sqrt {5}}}{2}}\end{aligned}}$ Note that the domain is all real line, so we don't need to check whether the points lie in the domain. Critical points: $\color {blue}{x={\dfrac {-1+{\sqrt {5}}}{2}},\ x={\dfrac {-1-{\sqrt {5}}}{2}}}$