Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 11 (c)

MATH 180 December 2017

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Question 11 (c)
The function $$P(x)={\frac {1}{4}}xe^{-x/2},\quad x\geq 0,$$ is an example of a Gamma distribution, a function that may be used to describe certain probabilities. (c) The derivative of ${\textstyle P(x)}$ is $$P'(x)=-{\frac {1}{8}}e^{-x/2}(x-2).$$ Identify all the intervals where ${\textstyle P(x)}$ is increasing, and all the intervals where ${\textstyle P(x)}$ 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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that a function is increasing if its derivative is positive, and decreasing if its derivative is negative.

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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. By the Hint, we need to look at the sign of ${\textstyle P'(x)}$. Since the fraction ${\textstyle {\dfrac {1}{8}}}$ and the exponential function ${\textstyle e^{-x/2}}$ are positive, the sign of ${\textstyle P'(x)}$ is the same as the sign of $$-(x-2).$$ For ${\textstyle x<2}$, that is for ${\textstyle x-2<0}$, we have that ${\textstyle -(x-2)>0}$. So ${\textstyle P'(x)>0}$. In other words, ${\textstyle P(x)}$ is increasing on the interval ${\textstyle [0,2)}$. Similarly, for ${\textstyle x>2}$ we have ${\textstyle x-2>0}$ and hence ${\textstyle -(x-2)<0}$. So ${\textstyle P'(x)<0}$ and ${\textstyle P(x)}$ is decreasing on the interval ${\textstyle (2,\infty )}$. Answer: ${\textstyle P(x)}$ is ${\textstyle {\color {blue}{\mbox{increasing on }}[0,2)}}$ and ${\textstyle {\color {blue}{\mbox{decreasing on }}(2,\infty )}}$.