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

Where $f^{″}(x)>0$, $f(x)$ is concave up; where $f^{″}(x)<0$, $f(x)$ is concave down. From part (a), we know that

$\begin{array}{rl}{f}^{\prime }(x)& =18(x^{-\frac{5}{7}}-x^{\frac{2}{7}})\\ f^{″}(x)& =18(-\frac{5}{7}\cdot x^{-\frac{5}{7}-1}-\frac{2}{7}\cdot x^{\frac{2}{7}-1})\\ & =18(-\frac{5}{7}{x}^{-\frac{12}{7}}-\frac{2}{7}{x}^{-\frac{5}{7}})\\ & =\frac{18}{7}{x}^{-\frac{12}{7}}(-5-2x)\end{array}$

Setting $f^{″}(x)=0$ and applying the zero product property:

$\begin{array}{rl}\frac{18}{7}{x}^{-\frac{12}{7}}(-5-2x)& =0\\ \frac{18}{7}{x}^{-\frac{12}{7}}=0-5-2x& =0\\ -5-2x& =0\\ x& =-\frac{5}{2}\end{array}$

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

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

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

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