Science:Math Exam Resources/Courses/MATH100/December 2013/Question 08 (c)/Solution 1

Where $\displaystyle f''(x)>0$, $\displaystyle f(x)$ is concave up; where $\displaystyle f''(x)<0$, $\displaystyle f(x)$ is concave down. From part (a), we know that

${\begin{aligned}f'(x)&=18\left(x^{-{\frac {5}{7}}}-x^{\frac {2}{7}}\right)\\f''(x)&=18\left(-{\tfrac {5}{7}}\cdot x^{-{\frac {5}{7}}-1}-{\tfrac {2}{7}}\cdot x^{{\frac {2}{7}}-1}\right)\\&=18\left(-{\tfrac {5}{7}}x^{-{\frac {12}{7}}}-{\tfrac {2}{7}}x^{-{\frac {5}{7}}}\right)\\&={\tfrac {18}{7}}x^{-{\frac {12}{7}}}(-5-2x)\end{aligned}}$

Setting $\displaystyle f''(x)=0$ and applying the zero product property:

${\begin{aligned}{\tfrac {18}{7}}x^{-{\frac {12}{7}}}(-5-2x)&=0\\{\color {red}{\tfrac {18}{7}}x^{-{\frac {12}{7}}}}=0\quad -5-2x&=0\\-5-2x&=0\\x&=-{\frac {5}{2}}\end{aligned}}$

Note that ${\tfrac {18}{7}}x^{-{\frac {12}{7}}}$ is never equal to zero on the interval $(-\infty ,\infty )$. However, ${\tfrac {18}{7}}x^{-{\frac {12}{7}}}$ is undefined at $x=0$, so $\displaystyle f''(x)$ does not exist at $x=0$.

Therefore, we note that $\displaystyle f''(x)$ may change sign at $\displaystyle x=0$ and $\displaystyle x=-{\tfrac {5}{2}}$. We can construct a sign table to organize our calculations:

	$\displaystyle (-\infty ,-{\tfrac {5}{2}})$	$\displaystyle (-{\tfrac {5}{2}},0)$	$\displaystyle (0,\infty )$
${\tfrac {18}{7}}x^{-{\frac {12}{7}}}$	$\displaystyle +$	$\displaystyle +$	$\displaystyle +$
$\displaystyle -5-2x$	$\displaystyle +$	$\displaystyle -$	$\displaystyle -$
$f''(x)=({\tfrac {18}{7}}x^{-{\frac {12}{7}}})(-5-2x)$	$\displaystyle +$	$\displaystyle -$	$\displaystyle -$

From this table we observe that $\displaystyle f(x)$ is concave up on ${\color {blue}x\in (-\infty ,-{\tfrac {5}{2}})}$ and concave down on ${\color {Orange}x\in (-{\tfrac {5}{2}},0)\cup (0,\infty )}$.