Chapter 4.3, Problem 20E

(a)

To determine

To find:The critical number of $f\left(x\right)=x^4{\left(x-1\right)}^3$

Explanation of Solution

Calculate the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=4x^3{\left(x-1\right)}^3+x^4\left(3\right){\left(x-1\right)}^2\\ =x^3{\left(x-1\right)}^2\left[4\left(x-1\right)+3x\right]\\ =x^3{\left(x-1\right)}^2\left(7x-4\right)\end{array}$

Equate the derivative to zero.

  $\begin{array}{c}{x}^3{\left(x-1\right)}^2\left(7x-4\right)=0\\ x=0,1,\frac{4}{7}\end{array}$

Therefore, the critical points are $0,1,\frac{4}{7}$ .

(b)

To determine

To find :The behaviour of the critical points by second derivative test.

Explanation of Solution

Calculate the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=4x^3{\left(x-1\right)}^3+x^4\left(3\right){\left(x-1\right)}^2\\ =x^3{\left(x-1\right)}^2\left[4\left(x-1\right)+3x\right]\\ =x^3{\left(x-1\right)}^2\left(7x-4\right)\end{array}$

Calculate the second derivative.

  $\begin{array}{c}{f}^{″}\left(x\right)=3x^2{\left(x-1\right)}^2\left(7x-4\right)+x^3\left[2\left(x-1\right)\right]\left(7x-4\right)+x^3{\left(x-1\right)}^{2}7\\ =x^2\left(x-1\right)\left[3\left(x-1\right)\left(7x-4\right)+2x\left(7x-4\right)+7x\left(x-1\right)\right]\end{array}$

At $x=0$

  $\begin{array}{c}{f}^{″}\left(0\right)=0^2\left(0-1\right)\left[3\left(0-1\right)\left(7\left(0\right)-4\right)+2\left(0\right)\left(7\left(0\right)-4\right)+7\left(0\right)\left(\left(0\right)-1\right)\right]\\ =0\end{array}$

At $x=1$

  $\begin{array}{c}{f}^{″}\left(1\right)=1^2\left(1-1\right)\left[3\left(1-1\right)\left(7\left(1\right)-4\right)+2\left(1\right)\left(7\left(1\right)-4\right)+7\left(1\right)\left(\left(1\right)-1\right)\right]\\ =0\end{array}$

At $x=\frac{4}{7}$

  $\begin{array}{c}{f}^{″}\left(\frac{4}{7}\right)={\frac{4}{7}}^2\left(\frac{4}{7}-1\right)\left[3\left(\frac{4}{7}-1\right)\left(7\left(\frac{4}{7}\right)-4\right)+2\left(\frac{4}{7}\right)\left(7\left(\frac{4}{7}\right)-4\right)+7\left(\frac{4}{7}\right)\left(\left(\frac{4}{7}\right)-1\right)\right]\\ =4{\left(\frac{4}{7}\right)}^2{\left(\frac{3}{7}\right)}^2\\ >0\end{array}$

The second derivative does not tell any behaviour of the function at $x=0,1$

Since, $f^{″}\left(\frac{4}{7}\right)>0$ , so the function attains local minimum at $x=\frac{4}{7}$

(c)

To determine

To find :The behaviour of the critical points by first derivative test.

Explanation of Solution

Calculate the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=4x^3{\left(x-1\right)}^3+x^4\left(3\right){\left(x-1\right)}^2\\ =x^3{\left(x-1\right)}^2\left[4\left(x-1\right)+3x\right]\\ =x^3{\left(x-1\right)}^2\left(7x-4\right)\end{array}$

The four regions are $\left(-\infty ,0\right),\left(0,\frac{4}{7}\right),\left(\frac{4}{7},1\right)\text{and}\left(1,\infty \right)$ .

The function has local maxima at $x-0$ and local minima at $x=\frac{4}{7}$ but neither at $x=1$ .