Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)

MATH110 April 2016

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Question 05 (f)
Let $f(x)={\frac {x-1}{x^{2}}}.$ (f) Find all local maximum and minimum values of $f$, if they exist.

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 first derivative test for local extremes; Let $f$ be a continuous function with $f'(a)=0$ or $f'(a)$ is undefined. (i) If $f'$(left of $a$) $>0$ and $f'$(right of $a$) $<0$, then $(a,f(a))$ is a local maximum (i) If $f'$(left of $a$) $<0$ and $f'$(right of $a$) $>0$, then $(a,f(a))$ is a local minimum.

Hint 2
(For the alternative solution) Recall the second derivative test for extremes: (1) If $f'(x)=0$ and ${\textstyle f''(x)<0}$, then ${\textstyle f}$ has a local maximum at ${\textstyle x}$; (2) If $f'(x)=0$ and ${\textstyle f''(x>0)}$, then ${\textstyle f}$ has a local minimum at ${\textstyle x}$;

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Solution 1

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 the part (a) and (d) of this question, we know that the domain of ${\textstyle f}$ is ${\textstyle \{x\neq 0\}}$ and the derivative of ${\textstyle f}$ is $$f'(x)={\frac {2-x}{x^{3}}}.$$ Then, we can easily see that the only critical point of ${\textstyle f}$ is ${\textstyle x=2}$ (because ${\textstyle f'(2)=0}$. Note that ${\textstyle x=0}$ is excluded because it is not in the domain.) Since the local extreme(s) only occur at the critical point(s), $x=2$ is the only candidate for the local extreme. In the part (e), we observe that ${\textstyle f}$ is increasing on ${\textstyle (0,2)}$, while it is decreasing on ${\textstyle (2,\infty )}$, by the first derivative test in the Hint, the point ${\textstyle \color {blue}(2,1/4)}$ is the local maximum.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (Alternative solution) From the solution 1, we know that the only critical point is ${\textstyle 2}$. Now let’s calculate the second derivative $$f''(x)=(f'(x))'=({\frac {2-x}{x^{3}}})'={\frac {(2-x)'x^{3}-(x^{3})'(2-x)}{x^{6}}}={\frac {2x-6}{x^{4}}}.$$ At ${\textstyle x=2}$, ${\textstyle f''(2)={\frac {2\times 2-6}{2^{4}}}=-{\frac {1}{8}}<0}$, so by the second derivative test in the Hint 2, it attains local maximum at this point. In all, it has no local minimum, but has a local maximum at ${\textstyle x=2}$ and ${\textstyle f(2)={\frac {2-1}{x^{2}}}={\frac {1}{4}}}$.