Chapter 5, Problem 9PT

a.

To determine

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

Explanation of Solution

Given information:

The equation, $g\left(x\right)=x^3+4x^2-3x+1$ .

Graph:

The graph of the quadratic surface $g\left(x\right)=x^3+4x^2-3x+1$ can be sketched using the traces,

For the plane $x=-4.686$

, the trace of the surfaces is:

  $\begin{array}{l}g\left(x\right)=x^3+4x^2-3x+1\\ g\left(-4\right)=-{4.686}^3+4{\left(-4.686\right)}^2-3\left(-4.686\right)+1\\ g\left(-4\right)=0\end{array}$

For the plane $x=-3$ , the trace of the surfaces is,

  $\begin{array}{l}g\left(x\right)=x^3+4x^2-3x+1\\ g\left(-3\right)=-3^3+4{\left(-3\right)}^2-3\left(-3\right)+1\\ g\left(-3\right)=-27+36+9+1\\ g\left(-3\right)=19\end{array}$

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

  $\begin{array}{l}g\left(x\right)=x^3+4x^2-3x+1\\ g\left(0\right)=0^3+4{\left(0\right)}^2-3\left(0\right)+1\\ g\left(0\right)=1\end{array}$

For the plane $x=0.333$ , the trace of the surfaces is,

  $\begin{array}{l}g\left(x\right)=x^3+4x^2-3x+1\\ g\left(0.333\right)=-{0.333}^3+4{\left(0.333\right)}^2-3\left(0.333\right)+1\\ g\left(0.333\right)=0.481\end{array}$

The values are shown in the table below:

  $\begin{array}{cc}x& g\left(x\right)\\ -4.686& 0\\ -3& 19\\ 0& 1\\ 0.333& 0.481\end{array}$

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 $0\text{ and 0}\text{.333,}$ in positive section of the graph.

It intersects the $x-\text{axis}$ at $\text{-4}\text{.686 and -3}$ in negative section of the graph.

It intersects the $y-\text{axis}$ at $\text{0 and 1,0}\text{.481,19}$ all are in positive section of the graph.

The graph increasing from $\text{-4}\text{.686,0 to -3,19}$ and decreasing from $\text{ -3,19 to 0}\text{.333,0}\text{.481}$ .

By the table observation the value of $x$ is maximum at the point $x=-3$ .

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

Therefore, the quadratic surface $g\left(x\right)=x^3+4x^2-3x+1$ 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 9PT

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

Explanation of Solution

Given information:

The expression “ $g\left(x\right)=x^3+4x^2-3x+1$ .”

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

Calculate:

Consider the provided statement “ $g\left(x\right)=x^3+4x^2-3x+1$ .”

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:

  $\begin{array}{cc}x& g\left(x\right)\\ -4.686& 0\\ -3& 19\\ 0& 1\\ 0.333& 0.481\end{array}$

There is only the change of sign which cuts the $x$ -axis there is a real zero between $x=-5\text{ to }x=-4$

Thus, the zeros are between “ $x=-5\text{ to }x=-4$ .”

c.

To determine

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

Answer to Problem 9PT

The relative maxima and minima is:

  $\begin{array}{l}\text{maxima},x=3\\ \text{minima},x=0\end{array}$ .

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:

  $\begin{array}{cc}x& g\left(x\right)\\ -4.686& 0\\ -3& 19\\ 0& 1\\ 0.333& 0.481\end{array}$

See in this table in which value of $x$ the $g\left(x\right)$

has a maximum value:

By the table observation the value of $x$ is maximum at the point $x=-3$ .

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

So, the relative maxima near $x=-3$ , and the relative minima is near by $x=0$ .

Thus, the relative maxima and minima is:

  $\begin{array}{l}\text{maxima},x=3\\ \text{minima},x=0\end{array}$ .