Chapter 4, Problem 17RE

$\text{(a)}$

To determine

To find: The value of local maximum.

Answer to Problem 17RE

The answer: no local maximum

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

  $\begin{array}{c}{y}^{″}=6\left(3x^2-6x\right)\\ =18x(x-2)\\ y^{″}(-1)=54>0\text{ concave up }\\ x=-1\text{ is a local minimum}\end{array}$

Then,

  $y^{″}(2)=0$

Intermediate

  $y^{\prime }(0)=24>0$

Increasing

  $y^{\prime }(3)=24>0$

Increasing.

Therefore, the required no local maximum of the given function.

$\text{(b)}$

To determine

To find: The value of local minimum.

Answer to Problem 17RE

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

  $\begin{array}{c}{y}^{″}=6\left(3x^2-6x\right)\\ =18x(x-2)\\ y^{″}(-1)=54>0\text{ concave up }\\ x=-1\text{ is a local minimum}\end{array}$

Then,

  $y^{″}(2)=0$

Intermediate

  $y^{\prime }(0)=24>0$

Increasing

  $y^{\prime }(3)=24>0$

Increasing.

Therefore, the required local minimum at $x=-1$ .

$(c)$

To determine

To find: The value of inflection point.

Answer to Problem 17RE

The answer: at $x_1=0$ and

At $x_2=2$ .

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{f}^{\prime }(x)=6(x+1)\left(x^2-4x+4\right)\\ =6\left(x^3-4x^2+4x+x^2-4x+4\right)\\ =6\left(x^3-3x^2+4\right)\\ f^{″}(x)=6{\left(x^3-3x^2+4\right)}^{\prime }\\ =6\left(3x^2-6x\right)\\ =18x(x-2)\end{array}$

Given that linear functions change their sign at their zeros, each of these numbers represent the values at which an inflection takes place.

Therefore, the required inflection points are at $x_1=0$ and

At $x_2=2$ .