Single Variable Calculus Questions and Answers

Start Your Free Trial

Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 42

Expert Answers info

eNotes eNotes educator | Certified Educator

calendarEducator since 2007

write13,548 answers

starTop subjects are Math, Literature, and Science

Sketch the curve of $y = x ^3 - 3a^2x+2a^3$; where $a$ is a positive constant.

If $y = x^3 - 3a^2 x + 2a^3$, then

$ \begin{equation} \begin{aligned} y' &= 3x^2 - 3a^2\\ y'' &= 6x \end{aligned} \end{equation} $

Solving for critical numbers, when $y'=0$

$ \begin{equation} \begin{aligned} 0 & = 3x^2 - 3a^2\\ 3x^2 &= 3a^2\\ x &= \sqrt{a^2}\\ x &= \pm a \end{aligned} \end{equation} $

Hence, we can divide the interval by;

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < -a & + & \text{increasing on } (-\infty,-a)\\ \hline\\ -a < x < a & - & \text{decreasing on } (-a,a)\\ \hline\\ x > a & + & \text{increasing on } (a,\infty)\\ \hline \end{array} $

Solving for the inflection point, when $y'' = 0$

$ \begin{equation} \begin{aligned} y'' = 0 &= 6x\\ x &= 0& \end{aligned} \end{equation} $

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < 0 & + & \text{Upward } \\ \hline\\ x > 0 & - & \text{Downward }\\ \hline \end{array} $

For the coordinates of the graph,

$ \begin{equation} \begin{aligned} \text{when } x &= a, &&& \text{when } x &= -a,\\ \\ f(a) &= a^3 - 3a^2(a) + 2a^3 &&& f(-a) &= (-a)^3 - 3a^2 (-a) + 2a^3\\ \\ f(a) &= 0 \text{ (Local Minimum)} &&& f(-a) &= 4a^3 \text{ (Local Maximum)}\\ \\ \text{when } x &= 0,\\ \\ f(0) &= (0)^3 - 3a^2 (0) + 2a^3\\ \\ f(0) &= 2a^3 \end{aligned} \end{equation} $

Therefore, we can illustrate the function as