Chapter 5.3, Problem 39E

a.

To determine

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

Answer to Problem 39E

There is no maximum.

Explanation of Solution

Given information :

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

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 $y\text{'}=0$ :

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

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

  

From the first derivative test the sign of $f\text{'}$ is negative of $x<2$ and positive for $x>2$ .

Since, sign of $f\text{'}$ from negative to positive. Therefore, there is no maximum .

Hence,

There is no maximum.

b.

To determine

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

Answer to Problem 39E

There is local minimum at $x=2$ .

Explanation of Solution

Given information :

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

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 $y\text{'}=0$ :

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

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

  

From the first derivative test the sign of $f\text{'}$ is negative of $x<2$ and positive for $x>2$ .

Therefore, there is local minimum at $x=2$

Hence,

There is local minimum at $x=2$ .

c.

To determine

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

Answer to Problem 39E

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

Explanation of Solution

Given information :

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

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:

  $\begin{array}{l}y\text{'}\text{'}={\left[{\left(x-1\right)}^2\right]}^{\text{'}}\left(x-2\right)+{\left(x-1\right)}^2\left(x-2\right)\text{'}\\ y\text{'}\text{'}=2\left(x-1\right)\left(x-2\right)+{\left(x-1\right)}^2\\ y\text{'}\text{'}=2x^2-6x+4+x^2-2x+1\\ y\text{'}\text{'}=3x^2-8x+5\\ y\text{'}\text{'}=3x^2-\left(3+5\right)x+5\\ y\text{'}\text{'}=3x^2-3x-5x+5\\ y\text{'}\text{'}=3x\left(x-1\right)-5\left(x-1\right)\\ y\text{'}\text{'}=\left(x-1\right)\left(3x-5\right)\end{array}$

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

  $\begin{array}{l}\left(x-1\right)\left(3x-5\right)=0\\ x-1=0\text{ or }3x-5=0\\ x=1\text{ or }3x=5\\ x-1=0\text{ or }3x-5=0\\ x=1\text{ or }x=\frac{5}{3}\end{array}$

Hence,

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