Science:Math Exam Resources/Courses/MATH104/December 2015/Question 10 (b)

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MATH104 December 2015

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Question 10 (b)
Let $f(x)=x^{2}+x-{\frac {1}{x}}-1$ . Its derivative satisfies $f'(x)=2x+1+{\frac {1}{x^{2}}}={\frac {2x^{3}+x^{2}+1}{x^{2}}}={\frac {(x+1)(2x^{2}-x+1)}{x^{2}}}$ . (b) Find the intervals on which $f$ is increasing or 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 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
Recall that a (differentiable) function $f(x)$ is increasing on an interval if $f'(x)\geq 0$ on that interval. Similarly, $f(x)$ is decreasing on an interval if $f'(x)\leq 0$ on that interval.

Hint 2
Since we are only interested in the sign of $f'$ and not its particular values, one of the given expressions for the derivative may be easier to work with than the others...

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Solution

To determine the intervals of increase/decrease of $f$ , we consider the sign of

f'(x)={\frac {(x+1)(2x^{2}-x+1)}{x^{2}}}\,.

Now $f'$ may change sign at the zeroes of the numerator (which are the zeroes of $f'$ itself; in this case $x=-1$ ) or the denominator (in this case $x=0$ ). Let us therefore tabulate our results as follows:

Interval:	$(-\infty ,-1)$	$(-1,0)$	$(0,\infty )$
Sign of $x+1$
Sign of $2x^{2}-x+1$
Sign of $x^{2}$
Sign of $f'(x)$

Considering the signs of each of the factors, we have

$2x^{2}-x+1{\color {OrangeRed}>0}$ for all $x$ (why?)
$x+1{\color {OrangeRed}>0}$ when $x>-1,$ while $x+1{\color {OliveGreen}<0}$ when $x<-1$
$x^{2}{\color {OrangeRed}>0}$ when $x>0$ and when $x<0$

Hence we can complete our table:

Interval:	$(-\infty ,-1)$	$(-1,0)$	$(0,\infty )$
Sign of $x+1$	${\color {OliveGreen}-}$	${\color {OrangeRed}+}$	${\color {OrangeRed}+}$
Sign of $2x^{2}-x+1$	${\color {OrangeRed}+}$	${\color {OrangeRed}+}$	${\color {OrangeRed}+}$
Sign of $x^{2}$	${\color {OrangeRed}+}$	${\color {OrangeRed}+}$	${\color {OrangeRed}+}$
Sign of $f'(x)$	${\tfrac {{\color {OliveGreen}-}\cdot {\color {OrangeRed}+}}{\color {OrangeRed}+}}={\color {OliveGreen}-}$	${\tfrac {{\color {OrangeRed}+}\cdot {\color {OrangeRed}+}}{\color {OrangeRed}+}}={\color {OrangeRed}+}$	${\tfrac {{\color {OrangeRed}+}\cdot {\color {OrangeRed}+}}{\color {OrangeRed}+}}={\color {OrangeRed}+}$

Since a (differentiable) function $f(x)$ is increasing on an interval if $f'(x)\geq 0$ on that interval (and similarly for decreasing), we have that $f(x)$ is increasing on ${\color {blue}[-1,\infty )\setminus \{0\}}$ and decreasing on ${\color {blue}(-\infty ,-1]}.$ Note that we have excluded $0$ from the interval of increase as $f$ is undefined at $x=0.$