MATH100 December 2013
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Question 08 (c)

FullSolution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise.
Let $\displaystyle y=f(x)=63x^{2/7}14x^{9/7}$. Determine the intervals where $\displaystyle f(x)$ is concave up, and the intervals where $\displaystyle f(x)$ is concave down.

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Solution

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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
${\begin{aligned}f'(x)&=18\left(x^{{\frac {5}{7}}}x^{\frac {2}{7}}\right)\\f''(x)&=18\left({\tfrac {5}{7}}\cdot x^{{\frac {5}{7}}1}{\tfrac {2}{7}}\cdot x^{{\frac {2}{7}}1}\right)\\&=18\left({\tfrac {5}{7}}x^{{\frac {12}{7}}}{\tfrac {2}{7}}x^{{\frac {5}{7}}}\right)\\&={\tfrac {18}{7}}x^{{\frac {12}{7}}}(52x)\end{aligned}}$
Setting $\displaystyle f''(x)=0$ and applying the zero product property:
${\begin{aligned}{\tfrac {18}{7}}x^{{\frac {12}{7}}}(52x)&=0\\{\color {red}{\tfrac {18}{7}}x^{{\frac {12}{7}}}}=0\quad 52x&=0\\52x&=0\\x&={\frac {5}{2}}\end{aligned}}$
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:

$\displaystyle (\infty ,{\tfrac {5}{2}})$ 
$\displaystyle ({\tfrac {5}{2}},0)$ 
$\displaystyle (0,\infty )$

${\tfrac {18}{7}}x^{{\frac {12}{7}}}$ 
$\displaystyle +$ 
$\displaystyle +$ 
$\displaystyle +$

$\displaystyle 52x$ 
$\displaystyle +$ 
$\displaystyle $ 
$\displaystyle $

$f''(x)=({\tfrac {18}{7}}x^{{\frac {12}{7}}})(52x)$ 
$\displaystyle +$ 
$\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 ${\color {Orange}x\in ({\tfrac {5}{2}},0)\cup (0,\infty )}$.

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