Chapter 6.8, Problem 10PT

a.

To determine

To graph: By making a table values graph the each function.

Explanation of Solution

Given information:

The equation, $h\left(x\right)=x^4-4x^3-3x^2+6x+2$ .

Graph:

The graph of the quadratic surface $h\left(x\right)=x^4-4x^3-3x^2+6x+2$ can be sketched using the traces,

For the plane $x=-2$ , the trace of the surfaces is:

  $\begin{array}{l}h\left(x\right)=x^4-4x^3-3x^2+6x+2\\ h\left(-2\right)=-2^4-4{\left(-2\right)}^3-3{\left(-2\right)}^2+6\left(-2\right)+2\\ h\left(-2\right)=16+32-12-12+2\\ h\left(-2\right)=26\end{array}$

For the plane $x=-1$ , the trace of the surfaces is,

  $\begin{array}{l}h\left(x\right)=x^4-4x^3-3x^2+6x+2\\ h\left(-1\right)=-1^4-4{\left(-1\right)}^3-3{\left(-1\right)}^2+6\left(-1\right)+2\\ h\left(-1\right)=1+4-3-6+2\\ h\left(-1\right)=-2\end{array}$

For the plane $x=0$ , the trace of the surfaces is,

  $\begin{array}{l}h\left(x\right)=x^4-4x^3-3x^2+6x+2\\ h\left(0\right)=0^4-4{\left(0\right)}^3-3{\left(0\right)}^2+6\left(0\right)+2\\ h\left(0\right)=2\end{array}$

For the plane $x=1$ , the trace of the surfaces is,

  $\begin{array}{l}h\left(x\right)=x^4-4x^3-3x^2+6x+2\\ h\left(1\right)=1^4-4{\left(1\right)}^3-3{\left(1\right)}^2+6\left(1\right)+2\\ h\left(1\right)=1-4-3+6+2\\ h\left(1\right)=2\end{array}$

For the plane $x=2$ , the trace of the surfaces is:

  $\begin{array}{l}h\left(x\right)=x^4-4x^3-3x^2+6x+2\\ h\left(2\right)=2^4-4{\left(2\right)}^3-3{\left(2\right)}^2+6\left(2\right)+2\\ h\left(2\right)=16-32-12+12+2\\ h\left(2\right)=-14\end{array}$

The values are shown in the table below:

  $\begin{array}{cc}x& h\left(x\right)\\ -2& 26\\ -1& -2\\ 0& 2\\ 1& 2\\ 2& -14\end{array}$

Then the graph of these values are shown below:

  

Interpretation:

The general equations of all quadratic surfaces along with the description of their traces is provided below,

The graph is a wave formed graph.

It intersects the $x-\text{axis}$ at $\text{1 and 2}$ in positive section of the graph.

It intersects the $x-\text{axis}$ at $\text{-1 and -2}$ in negative section of the graph.

By the table observation the value of $x$ is maximum at the point $x=-2$ .

And the value of $x$ is minimum at the point $x=2$ .

Therefore, the quadratic surface $h\left(x\right)=x^4-4x^3-3x^2+6x+2$ is wave formed graph with axis as x -axis.

b.

To determine

To calculate: The consecutive integers value of $x$ between which each zero is located.

Answer to Problem 10PT

The integer value of $x$ between which each zero is located “ $x=-5\text{ to }x=-4$ .”

Explanation of Solution

Given information:

The expression “ $h\left(x\right)=x^4-4x^3-3x^2+6x+2$ .”

The integer value of $x$ between which each zero is located “ $x=-1\text{ to }x=0$ .”

Calculate:

Consider the provided statement “ $h\left(x\right)=x^4-4x^3-3x^2+6x+2$ .”

When take out roots of the equation, check for the value of $x$ becomes zero at which points.

Now observe this table to get the result:

  $\begin{array}{cc}x& h\left(x\right)\\ -2& 26\\ -1& -2\\ 0& 2\\ 1& 2\\ 2& -14\end{array}$

There is only the change of sign which cuts the $x$ -axis there is a real zero between $x=-1\text{ to }x=0$

Thus, the zeros are between “ $x=-1\text{ to }x=0$ .”

c.

To determine

To calculate: The $x$ -coordinates at which the relative maxima and minima occur.

Answer to Problem 10PT

The relative maxima and minima is:

  $\begin{array}{l}\text{maxima},x=0.5\\ \text{minima},x=2\end{array}$ .

Explanation of Solution

Given information:

The $x$ -coordinates at which the relative maxima and minima occur.

Calculate:

Consider the provided statement “The $x$ -coordinates at which the relative maxima and minima occur”

Here see in the graph points or in the table of values:

  $\begin{array}{cc}x& h\left(x\right)\\ -2& 26\\ -1& -2\\ 0& 2\\ 1& 2\\ 2& -14\end{array}$

See in this table in which value of $x$ the $h\left(x\right)$ has a maximum value:

By the table observation the value of $x$ is maximum at the point $x=-2$ .

And the value of $x$ is minimum at the point $x=0$ .

So, the relative maxima near $x=-2$ , and the relative minima is near by $x=0$ .

Thus, the relative maxima and minima is:

  $\begin{array}{l}\text{maxima},x=-2\\ \text{minima},x=0\end{array}$ .