Chapter 5.3, Problem 40E

a.

To determine

To find : The point at which f has a local maximum.

Answer to Problem 40E

The local maximum is at $x=2$ .

Explanation of Solution

Given information :

The given function is $y\text{'}={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ .

Calculation :

For the continuous function $f\left(x\right)$ .

At the critical point $c$ :

1. If $f$ changes sign from positive to negative at $c$ ( $f\text{'}>c$ for $x<c$ and $f\text{'}<0$ for $x>c$ ), then $f$ has a local maximum value at $c$ .
2. if $f$ changes sign from negative to positive at $c$ ( $f\text{'}<0$ for $x<c$ and $f\text{'}>c$ for $x>c$ ), then $f$ has a local minimum value at $c$ .
3. if $f$ does not changes its sign at $c$ ( $f\text{'}$ has same sign on both sides of $c$ ), then $f$ has no local extreme value at $c$ .

Find the critical value by setting $y\text{'}=0$ :

  $\begin{array}{l}{\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)=0\\ {\left(x-1\right)}^2=0\text{ or }x-2=0\text{ or }x-4=0\\ x-1=0\text{ or }x=2\text{ or }x=4\\ x=1\text{ or }x=2\text{ or }x=4\end{array}$

The critical point partition the x-axis into four intervals as shown below:

  

From the first derivative test the sign of $f\text{'}$ changes from positive to negative at $x=2$ .

Therefore, the local maximum is at $x=2$ .

Hence,

The local maximum is at $x=2$ .

b.

To determine

To find : The point at which f has a local minimum.

Answer to Problem 40E

The local minimum is at $x=4$ .

Explanation of Solution

Given information :

The given function is $y\text{'}={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ .

Calculation :

For the continuous function $f\left(x\right)$ .

At the critical point $c$ :

1. If $f$ changes sign from positive to negative at $c$ ( $f\text{'}>c$ for $x<c$ and $f\text{'}<0$ for $x>c$ ), then $f$ has a local maximum value at $c$ .
2. if $f$ changes sign from negative to positive at $c$ ( $f\text{'}<0$ for $x<c$ and $f\text{'}>c$ for $x>c$ ), then $f$ has a local minimum value at $c$ .
3. if $f$ does not changes its sign at $c$ ( $f\text{'}$ has same sign on both sides of $c$ ), then $f$ has no local extreme value at $c$ .

Find the critical value by setting $y\text{'}=0$ :

  $\begin{array}{l}{\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)=0\\ {\left(x-1\right)}^2=0\text{ or }x-2=0\text{ or }x-4=0\\ x-1=0\text{ or }x=2\text{ or }x=4\\ x=1\text{ or }x=2\text{ or }x=4\end{array}$

The critical point partition the x-axis into four intervals as shown below:

  

From the first derivative test the sign of $f\text{'}$ changes from negative to positive at $x=4$ .

Therefore, the local minimum is at $x=4$ .

Hence,

The local minimum is at $x=4$ .

c.

To determine

To find : The point at which f has a point of inflection.

Answer to Problem 40E

The point of inflection is $x=1$ and $x=\frac{5\pm \sqrt{3}}{2}$ .

Explanation of Solution

Given information :

The given function is $y\text{'}={\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)$ .

Calculation :

For the continuous function $f\left(x\right)$

The point where the graph of the function has a tangent line and where the concavity changes is a point of inflection, that is when $y\text{'}\text{'}=0$

Differentiate the given function by using product rule:

  $\begin{array}{l}y\text{'}\text{'}={\left[{\left(x-1\right)}^2\right]}^{\text{'}}\left(x-2\right)\left(x-4\right)+{\left(x-1\right)}^2\left(x-2\right)\text{'}\left(x-4\right)+{\left(x-1\right)}^2\left(x-2\right)\left(x-4\right)\text{'}\\ y\text{'}\text{'}=\left[2\left(x-1\right)\left(1\right)\right]\left(x-2\right)\left(x-4\right)+{\left(x-1\right)}^2\left(1\right)\left(x-4\right)+{\left(x-1\right)}^2\left(x-2\right)\left(1\right)\\ y\text{'}\text{'}=2\left(x-1\right)\left(x-2\right)\left(x-4\right)+{\left(x-1\right)}^2\left(x-4\right)+{\left(x-1\right)}^2\left(x-2\right)\\ y\text{'}\text{'}=\left(x-1\right)\left[2\left(x-2\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)+\left(x-1\right)\left(x-2\right)\right]\\ y\text{'}\text{'}=\left(x-1\right)\left[2x^2-12x+16+x^2-5x+4+x^2-3x+2\right]\\ y\text{'}\text{'}=\left(x-1\right)\left[4x^2-20x+22\right]\\ y\text{'}\text{'}=2\left(x-1\right)\left[2x^2-10x+11\right]\end{array}$

For point of inflection set $y\text{'}\text{'}=0$ and solve for x:

  $\begin{array}{l}2\left(x-1\right)\left(2x^2-10x+11\right)=0\\ \left(x-1\right)\left(2x^2-10x+11\right)=0\\ \left(x-1\right)=0\text{ or }\left(2x^2-10x+11\right)=0\\ x=1\text{ or }2x^2-10x+11=0\end{array}$

Solve the quadratic part:

  $\begin{array}{l}2x^2-10x+11=0\\ x=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\left(2\right)\left(11\right)}}{2\left(2\right)}\\ x=\frac{10\pm \sqrt{100-88}}{4}\\ x=\frac{10\pm \sqrt{12}}{4}\\ x=\frac{10\pm 2\sqrt{3}}{4}\\ x=\frac{5\pm \sqrt{3}}{2}\end{array}$

Hence,

The point of inflection is $x=1$ and $x=\frac{5\pm \sqrt{3}}{2}$ .