Science:Math Exam Resources/Courses/MATH100/December 2013/Question 08 (b)

MATH100 December 2013

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Question 08 (b)
Full-Solution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise. Let $\displaystyle y=f(x)=63x^{2/7}-14x^{9/7}$. Determine the intervals where $\displaystyle f(x)$ is increasing, and the intervals where $\displaystyle f(x)$ is decreasing. (The results from part (a) may be useful.)

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
What is the sign of $\displaystyle f'(x)$ when $\displaystyle f(x)$ is increasing? When $\displaystyle f(x)$ is decreasing?

Hint 2
We know from part (a) that $f'(x)={\frac {18(1-x)}{x^{\frac {5}{7}}}}$ and that $\displaystyle f'(x)=0$ when $\displaystyle x=1$. How can we use this to find where $\displaystyle f'(x)<0$ and where $\displaystyle f'(x)>0$?

Hint 3
Note also that $\displaystyle f'(x)$ has a vertical asymptote at $\displaystyle x=0$. What can we say about the signs of $\displaystyle f'(x)$ when $\displaystyle x<0$ and when $\displaystyle 0<x<1$?

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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. Where $\displaystyle f'(x)>0$, $\displaystyle f(x)$ is increasing; where $\displaystyle f'(x)<0$, $\displaystyle f(x)$ is decreasing. From part (a), we note that $\displaystyle f'(x)$ may change sign at its critical points $\displaystyle x=0$ and $\displaystyle x=1$. We can construct a sign table to organize our calculations: $\displaystyle (-\infty ,0)$ $\displaystyle (0,1)$ $\displaystyle (1,\infty )$ $18x^{-{\frac {5}{7}}}$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle 1-x$ $\displaystyle +$ $\displaystyle +$ $\displaystyle -$ $f'(x)=(18x^{-{\frac {5}{7}}})(1-x)$ $\displaystyle -$ $\displaystyle +$ $\displaystyle -$ From this table we observe that $\displaystyle f(x)$ is increasing on ${\color {blue}x\in (0,1)}$ and decreasing on ${\color {OliveGreen}x\in (-\infty ,0)\cup (1,\infty )}$.

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