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

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

• $\underset{x\to 0^{-}}{\lim }{f}^{\prime }(x)=-\mathrm{\infty }$ and $\underset{x\to 0^{+}}{\lim }{f}^{\prime }(x)=\mathrm{\infty }$, 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:

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