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

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 (d)
Consider the function $f(x)={\frac {x^{2}}{x^{2}-4}}.$ Its first and second derivatives are given by $f'(x)=-{\frac {8x}{(x^{2}-4)^{2}}},\quad f''(x)={\frac {8(3x^{2}+4)}{(x^{2}-4)^{3}}}$ (d) On which intervals is $f(x)$ concave up? On which intervals is $f(x)$ concave down?

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 second derivative on the intervals between the points found in part (b).

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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. We proceed as suggested by the hint. On the interval $\displaystyle (-\infty ,-2)$, we see that the second derivative at the point $\displaystyle x=-10$ is given by $\displaystyle f''(-10)={\frac {8(3(-10)^{2}+4)}{((-10)^{2}-4)^{3}}}>0$ and so the function is concave up on this interval. On the interval $\displaystyle (-2,2)$, we see that the second derivative at the point $\displaystyle x=0$ is given by $\displaystyle f''(0)={\frac {8(3(0)^{2}+4)}{((0)^{2}-4)^{3}}}<0$ and so the function is concave down on this interval. On the interval $\displaystyle (2,\infty )$, we see that the second derivative at the point $\displaystyle x=10$ is given by $\displaystyle f''(10)={\frac {8(3(10)^{2}+4)}{((10)^{2}-4)^{3}}}>0$ and so the function is concave up on this interval. To summarize, $f(x)$ is concave up on $(-\infty ,-2)\cup (2,\infty )$ and concave down on $(-2,2).$

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. We can also argue a bit more intelligently as follows. Look at the second derivative given by $f''(x)={\frac {8(3x^{2}+4)}{(x^{2}-4)^{3}}}$ Notice that the numerator of the second derivative is always positive and so the sign of the second derivative is given by the behaviour of the denominator, namely the polynomial $(x^{2}-4)^{3}$ Since this polynomial is to the power of 3, the sign of the cube of the polynomial is the same as the sign of the polynomial itself, given by $x^{2}-4$ Next, this polynomial is an upwards facing parabola with roots at $\pm 2$. Thus, it is negative between $-2<x<2$ and is positive on $(-\infty ,-2)\cup (2,\infty )$. Hence, $f(x)$ is concave up on $(-\infty ,-2)\cup (2,\infty )$ and concave down on $(-2,2).$

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