Science:Math Exam Resources/Courses/MATH104/December 2013/Question 02 (c)

MATH104 December 2013

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q2 (f)  •  Q2 (g)  •  Q3  •  Q4  •  Q5  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  

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Question 02 (c)
Consider the function $f(x)={\frac {x^{2}}{x^{2}-4}}.$ Its first and second derivatives are given by $f'(x)=-{\frac {8x}{(x^{2}-4)^{2}}},\quad f''(x)={\frac {8(3x^{2}+4)}{(x^{2}-4)^{3}}}$ (c) On which intervals is $f(x)$ increasing? On which intervals is $f(x)$ 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
Remember we need to check the sign of the derivative on the intervals between the critical points found in part (a).

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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. We proceed as suggested by the hint. On the interval $\displaystyle (-\infty ,-2)$, we see that the derivative at the point $\displaystyle x=-10$ is given by $\displaystyle f'(-10)={\frac {-8(-10)}{((-10)^{2}-4)^{2}}}>0$ and so the function is increasing on this interval. On the interval $\displaystyle (-2,0)$, we see that the derivative at the point $\displaystyle x=-1$ is given by $\displaystyle f'(-1)={\frac {-8(-1)}{((-1)^{2}-4)^{2}}}>0$ and so the function is increasing on this interval. On the interval $\displaystyle (0,2)$, we see that the derivative at the point $\displaystyle x=1$ is given by $\displaystyle f'(1)={\frac {-8(1)}{((1)^{2}-4)^{2}}}<0$ and so the function is decreasing on this interval. On the interval $\displaystyle (2,\infty )$, we see that the derivative at the point $\displaystyle x=10$ is given by $\displaystyle f'(10)={\frac {-8(10)}{((10)^{2}-4)^{2}}}<0$ and so the function is decreasing on this interval.

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