Science:Math Exam Resources/Courses/MATH104/December 2015/Question 10 (b)/Solution 1

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 $x=0.$