Chapter 4, Problem 18RE

$\text{(a)}$

To determine

To find: The value of local maximum.

Answer to Problem 18RE

The answer: local minimum at $x=-1$

Explanation of Solution

Given information:

The function is $y^{\prime }=6(x+1){(x-2)}^2$

Calculation:

Where, $y^{\prime }=0$

So,

  $\begin{array}{c}$ y^{\prime }=6(x+1){(x-2)}^2$$ =6\left(x^3-3x^2+4\right)$$ 0=6(x+1){(x-2)}^2$x=-1,2\end{array}$

And

  $\begin{array}{c}{y}^{″}=6(2x-1)\\ y^{″}(-1)=-18<0\\ x=-1\end{array}$

Is not a local minimum.

  $\begin{array}{c}{y}^{″}(2)=18>0\\ x=2\end{array}$

Is a local minimum.

Therefore, the required local maximum of the given function at $x=-1$ .

$\text{(b)}$

To determine

To find: The value of local minimum.

Answer to Problem 18RE

The answer: local minimum at $x=2$ .

Explanation of Solution

Given information:

The function is $y^{\prime }=6(x+1){(x-2)}^2$

Calculation:

Where, $y^{\prime }=0$

So,

  $\begin{array}{c}$ y^{\prime }=6(x+1){(x-2)}^2$$ =6\left(x^3-3x^2+4\right)$$ 0=6(x+1){(x-2)}^2$x=-1,2\end{array}$

And

  $\begin{array}{c}{y}^{″}=6(2x-1)\\ y^{″}(-1)=-18<0\\ x=-1\end{array}$

Is not a local minimum.

  $\begin{array}{c}{y}^{″}(2)=18>0\\ x=2\end{array}$

Is a local minimum.

Therefore, the required local minimum at $x=2$ .

$(c)$

To determine

To find: The value of inflection point.

Answer to Problem 18RE

The answer: at $x=0.5$

Explanation of Solution

Given information:

The function is $y^{\prime }=6(x+1){(x-2)}^2$

Calculation:

  $\begin{array}{c}{f}^{\prime }(x)=6\left(x^2-2x+x-2\right)\\ =6\left(x^2-x-2\right)\end{array}$

And

  $\begin{array}{c}{f}^{″}(x)=6(2x-1)\\ f^{″}(x)=0\\ \iff 2x-1=0\\ \iff x=0.5\end{array}$

Therefore, the required inflection point is $x=0.5$ .