Chapter 4.3, Problem 42E

To determine

To find: The member of the family of the curve, and sketch the curve by using method of this section.

Answer to Problem 42E

The all the members of family are concave downward on the interval $\left(-\infty ,0\right)$ , and concave upward on the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=x^3-3a^{2}x+2a^3$ .

Calculation:

Consider the function.

  $f\left(x\right)=x^3-3a^{2}x+2a^3$ .

First find the intervals on which the function $f$ is increasing or decreasing.

  $\left(a\right)$ If $f\text{'}\left(x\right)>0$ on an interval, then $f$ is increasing on that interval.

  $\left(b\right)$ If $f\text{'}\left(x\right)<0$ on an interval, then $f$ is decreasing on that interval.

Thus, calculate the derivative of the function.

  $\begin{array}{c}f\left(x\right)=\frac{d}{dx}\left(x^3-3a^{2}x+2a^3\right)\\ =\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(-3a^{2}x\right)+\frac{d}{dx}\left(2a^3\right)\\ =3x^2-3a^2\left(1\right)+2a^3\left(0\right)\\ =3x^2-3a^2\\ =3\left(x+a\right)\left(x-a\right)\end{array}$

Now, find the critical numbers of the function $f$ by equating $f\text{'}\left(x\right)$ to zeeo.

  $3\left(x+a\right)\left(x-a\right)$

  $x=-a$ or $x=a$ .

Thus the critical numbers are $-a,a$ .

Construct a chart by arranging our work.

Interval	$x+a$	$x-a$	$f\text{'}\left(x\right)$	$f$
$x<-a$	$+$	$-$	$+$	Increasing on $\left(-\infty ,-a\right)$
$-a<x<a$	$+$	$-$	$+$	Decreasing on $\left(-a,a\right)$
$x>a$	$+$	$+$	$+$	Increasing on $\left(a,\infty \right)$

Therefore the function $f\left(x\right)=x^2$ in $x$ is decreasing on $\left(-a,a\right)$ and increasing on $\left(-\infty ,-a\right)$ and $\left(a,\infty \right)$ .

To find the local maximum and local minimum values of $f$ .

Suppose that $c$ id a critical number of a continuous function $f$ .

  $\left(a\right)$ If $f\text{'}$ change from positive to negative at $c$ , then $f$ has local maximum at $c$ .

  $\left(b\right)$ If $f\text{'}$ change from negative to positive at $c$ , then $f$ has local minimum at $c$ .

  $\left(c\right)$ If $f\text{'}$ does not change sign at $c$ , then $f$ has no local maximum or minimum at $c$ .

From the above chart, $f\text{'}\left(x\right)$ change from positive to negative at $-a$ .

Thus, $f\left(x\right)=x^3-3a^{2}x+2a^3$ has minimum value.

  $\begin{array}{c}f\left(-a\right)={\left(-a\right)}^3-3a^2\left(-a\right)+2a^3\\ =-a^3+3a^3+2a^3\\ =4a^3\end{array}$

And $f\text{'}\left(x\right)$ change from negative to positive at $a$ .

Thus $f\left(x\right)=x^3-3a^{2}x+2a^3$ has minimum value.

  $\begin{array}{c}f\left(a\right)={\left(a\right)}^3-3a^2\left(a\right)+2a^3\\ =-a^3-3a^3+2a^3\\ =0\end{array}$

Therefore, the function $f$ has local maximum $\left(-a,4a^3\right)$ and it has local minimum $\left(a,0\right)$ .

Now calculate the derivative of the function $f\text{'}\left(x\right)=3x^2-3a^2$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{d}{dx}\left(3x^2-3a^2\right)\\ f\text{'}\left(x\right)=\frac{d}{dx}\left(3x^2\right)+\frac{d}{dx}\left(-3a^2\right)\\ =3\frac{d}{dx}\left(x^2\right)-3a^2\frac{d}{dx}\left(1\right)\\ =3\left(2x\right)-3a^2\left(0\right)\\ =6x-0\\ =0\end{array}$ 2

Thus, $f\text{'}\left(x\right)=6x$ .

Since, $f\text{'}\left(x\right)=6x=0$ when $x=0$ , so, divide the real line into intervals with this number as endpoint and compute the following chart.

Interval	$f\text{'}\left(x\right)=6x$	Concavity
$\left(-\infty ,0\right)$	$-$	Downward
$\left(0,\infty \right)$	$+$	upward

Graph for function $f\left(x\right)=x^3-3a^{2}x+2a^3$ by using the intervals of increasing and decreasing, local extreme value, the interval of concavity, and the infliction point as below.

  

Graph $\left(1\right)$

Graph of curve of the function $f\left(x\right)=x^3-3a^{2}x+2a^3$ by using substitute several values for the positive constant $a$ as below,

  

Graph $\left(2\right)$

Therefore, all the members of family are concave downward on the interval $\left(-\infty ,0\right)$ , and concave upward on the interval $\left(0,\infty \right)$ .