Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 15

MATH100 A December 2023

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Question 15
Let $f(x)=x+{\frac {4}{x}}$ be defined on the domain $\{x>0\}$. Find the $x$-values of all the extrema on that interval, indicating for each $x$-value whether there is a local maximum or a local minimum there.

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
By definition, a local extremum is a point $(c,f(c))$ such that $f'(c)=0$.

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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 extrema, we must solve the equation $f'(x)=0$. We have $f'(x)=1-{\frac {4}{x^{2}}}.$ Thus, we must find the values of $x$ that satisfy $1-{\frac {4}{x^{2}}}=0.$ Rearranging, the equation above is equivalent to $x^{2}=4,$ so the only $x$ that satisfies $f'(x)=0$ is $x=2$ (the domain of $f$ is the positive real axis). Thus, the only extremum is $(2,f(2))=(2,4)$. To find out if this is a local maximum or minimum, let us compute the second derivative and check for concavity. We have $f''(x)={\frac {8}{x^{3}}}$, so $f''(2)=1>0$, and thus the slope of $f$ is increasing at $x=2$. The graph of $f$ is then concave up and the extremum $(2,4)$ is a local minimum.