Science:Math Exam Resources/Courses/MATH104/December 2015/Question 02 (b)

MATH104 December 2015

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Question 02 (b)
Find the absolute maximum of $f(x)=\ln(x^{2}+x+1)$ over the interval $[-1,1].$

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 Fermat's theorem (which you may have encountered under a different name). It states that if $f$ is a differentiable function on some open interval $(a,b)$ and $c\in (a,b)$ is a local extremum of $f,$ then $f'(c)=0.$ In the context of this problem, this means that points in $(-1,1)$ where $f'$ is zero are possible locations of local maxima.

Hint 2
Recall that the extreme value theorem states that a continuous function $f$ defined on an interval $[a,b]$ attains an (absolute) maximum and an (absolute) minimum on this interval. In the context of this problem, this means that there is a $c\in [-1,1]$ such that $f(c)\geq f(x)$ for all $x\in [a,b].$ Make sure to consider the endpoints of the interval!

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We first find the derivative of $f$ and set it equal to zero to find $x$ for which $f(x)$ is possibly at a maximum: $f'(x)={\frac {1}{x^{2}+x+1}}\cdot (x^{2}+x+1)'={\frac {2x+1}{x^{2}+x+1}}$ ${\frac {2x+1}{x^{2}+x+1}}=0\Longleftrightarrow 2x+1=0\Longleftrightarrow x=-{\frac {1}{2}}\,.$ It turns out that $f(-{\tfrac {1}{2}})=\ln((-{\tfrac {1}{2}})^{2}+-{\tfrac {1}{2}}+1)={\color {OliveGreen}\ln {\tfrac {3}{4}}}.$ Now we check the endpoints: $f(-1)=\ln((-1)^{2}+(-1)+1)={\color {OliveGreen}\ln 1}=0,$ $f(1)=\ln(1^{2}+1+1)={\color {OliveGreen}\ln 3}\,.$ Finally, we have $\max \nolimits _{x\in [-1,1]}f(x)=\max\{{\color {OliveGreen}\ln {\tfrac {3}{4}}},{\color {OliveGreen}\ln 1},{\color {OliveGreen}\ln 3}\}.$ The greatest value is $\ln 3,$ since $\ln$ is a strictly increasing function ($\ln {\tfrac {3}{4}}<\ln 1<\ln 3$). Even if we did not know this, we could have observed that $\ln {\tfrac {3}{4}}<0,\ln 1=0,\ln 3>0$ to arrive at the same conclusion. Thus $\max \nolimits _{x\in [-1,1]}f(x)={\color {blue}\ln 3}.$