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

From parts (a), (b), and (c) we know the following about $f(x)=63x^{2/7}-14x^{9/7}$:

• The critical numbers of $f$ (where $f'$ is zero or does not exist) are $\displaystyle x=0,1$
• $f$ is increasing on $x\in (0,1)$ and decreasing on $x\in (-\infty ,0)\cup (1,\infty )$
• $f$ has a local minimum at $\displaystyle x=0$ and a local maximum at $\displaystyle x=1$ (note the change in the increase/decrease of $f$)
• $f$ is concave up on $x\in (-\infty ,-{\tfrac {5}{2}})$ and concave down on $x\in (-{\tfrac {5}{2}},0)\cup (0,\infty )$
• $f$ has an inflection point (changes concavity) at $x=-{\tfrac {5}{2}}$

• $\displaystyle f(x)=x^{\frac {2}{7}}(63-14x)$ and thus $f$ has x-intercepts at $x=0,{\tfrac {9}{2}}$

• $\lim _{x\to 0^{-}}f'(x)=-\infty$ and $\lim _{x\to 0^{+}}f'(x)=\infty$, thus the graph of $f$ is very steep near $x=0$

Plotting the points of interest gives the following:

Accounting for increase and decrease gives the following rough sketch:

Finally, incorporating concavity leads us to the final graph:

$\displaystyle f(x)=63x^{2/7}-14x^{9/7}$