Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (c)

MATH110 April 2016

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Question 02 (c)
Find the absolute minimum of $f(x)=\ln(x^{2}+x+1)$ on the interval $[-1,1]$. Make sure you justify your answer.

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
The absolute minimum of $f(x)$ on the interval $[-1,1]$ may be found at local minimum or two ends.

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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. To find the local minimum we need to compute the derivate of $f(x)$. Using the chain rule, $f(x)=g(h(x))$, where $g(x)=\ln x,h(x)=x^{2}+x+1$ we have $f'(x)={\frac {1}{x^{2}+x+1}}\cdot (2x+1)={\frac {2x+1}{x^{2}+x+1}}$ local maximum/minimum : $f'(x)={\frac {2x+1}{x^{2}+x+1}}=0\rightarrow x=-0.5\rightarrow f(-0.5)=\ln 0.75<0$ two ends: $f(1)=ln(3),f(-1)=ln1=0$ Therefore, the absolute minimum is when $x=-0.5$, it is $\ln 0.75$