Science:Math Exam Resources/Courses/MATH100/December 2015/Question 06 (c)

MATH100 December 2015

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Question 06 (c)
What are the critical points of the function $f(x)=e^{x^{3}-9x^{2}+15x-1}$?

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
Recall the definition of the critical point.

Hint 2
What is the differentiation rule for the function $e^{f(x)}$?

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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. Recall that $x$ is a critical point of $f$ if $f'(x)$ either vanishes or doesn't exist. First, we find the derivative of $f(x)=e^{x^{3}-9x^{2}+15x-1}$. Using the chain rule, we have $f'(x)=(e^{x^{3}-9x^{2}+15x-1})'=e^{x^{3}-9x^{2}+15x-1}(x^{3}-9x^{2}+15x-1)'=e^{x^{3}-9x^{2}+15x-1}(3x^{2}-18x+15)$. Since $f'$ is defined at any point $x\in \mathbb {R}$, it is enough to find a point $x$ such that $f'(x)=0$. Note that $e^{y}>0$ for any $y\in \mathbb {R}$, so that $e^{x^{3}-9x^{2}+15x-1}>0$ at any point $x\in \mathbb {R}$. Therefore, we have $f'(x)=e^{x^{3}-9x^{2}+15x-1}(3x^{2}-18x+15)=0\iff 3x^{2}-18x+15=3(x^{2}-6x+5)=3(x-1)(x-5)=0,$ and hence the critical points of $f$ are $\color {blue}{x=1,x=5}$.