Science:Math Exam Resources/Courses/MATH100/December 2012/Question 06 (a)

MATH100 December 2012

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Question 06 (a)

Full-Solution Problems. In questions 5-11, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. 6. Let $f(x)=2x^{3/5}+3x^{-2/5}$. Note that the domain of ƒ is the set of all nonzero real numbers; for example, $f(-1)=-2+3=1$. (a) Determine the interval(s) where $f(x)$ is increasing, and the interval(s) where $f(x)$ is decreasing.

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
Differentiate and check the sign of the derivative between critical points.

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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. Differentiating $\displaystyle f(x)=2x^{3/5}+3x^{-2/5}$ gives $\displaystyle f'(x)={\frac {6x^{-2/5}}{5}}+{\frac {-6x^{-7/5}}{5}}={\frac {6x^{-7/5}}{5}}(x-1)$. Thus the critical numbers are $x=1$ and $x=0$ (notice here our derivative and our function are both undefined). So we check the intervals $\displaystyle (-\infty ,0),(0,1),(1,\infty )$ individually. Case 1: $\displaystyle (-\infty ,0)$ We test that the derivative at a point in the interval, say $x=-1$, which gives a value of $12/5>0$. Thus the function is increasing on this interval. Case 2: $\displaystyle (0,1)$ We test that the derivative at a point in the interval, say $x=1/2$, which gives a value of $6\cdot 2^{7/5}/5(-1/2)<0$. Thus the function is decreasing on this interval. Case 3: $\displaystyle (1,\infty )$ We test that the derivative at a point in the interval, say $x=2$, which gives a value of $6\cdot 2^{-7/5}/5>0$. Thus the function is increasing on this interval. Summary Overall, the function in increasing on $(-\infty ,0)\cup (1,\infty )$ and decreasing on $\displaystyle (0,1)$.

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