Chapter 3.5, Problem 36E

a.

To determine

To find: the intervals where $f\left(x\right)$ is increasing.

Answer to Problem 36E

  $-\infty \le x\le 4$

Explanation of Solution

Given information:

A line is secant to the graph of $f\left(x\right)$ if it intersects the graph in at least two distinct points

  $f\left(x\right)=-{\left(x-4\right)}^2-3$

Calculation:

Plot the graph using a graphing calculator as look for the point of maxima in the graph. The graph of $f\left(x\right)$ is shown below.

  

The graph is increasing from $x=-\infty$ to $x=4$ . So, the interval for which $f\left(x\right)$ increases is $-\infty \le x\le 4$ .

b.

To determine

To find: slope of the secant line that passes through any two points contained in the interval evaluated in part a.

Answer to Problem 36E

5

Explanation of Solution

Given information:

A line is secant to the graph of $f\left(x\right)$ if it intersects the graph in at least two distinct points

  $f\left(x\right)=-{\left(x-4\right)}^2-3$

The increasing intervals evaluated in part a. are $-\infty \le x\le 4$ .

Calculation:

The two point selected from the increasing interval to find the slope are $\left(1,-12\right)$ and $\left(2,-7\right)$ .

Now put the points in the slope formula and evaluate m as follows:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-7-\left(-12\right)}{2-1}\\ =\frac{-7+12}{1}\\ =5\end{array}$

Thus, the slope of the secant line is 5.

c.

To determine

To make: a conjecture about the slope of any secant line that passes through any two points contained in the interval evaluated in part a.

Answer to Problem 36E

The slope of any line made on an interval where the graph is increasing will be positive.

Explanation of Solution

Given information:

A line is secant to the graph of $f\left(x\right)$ if it intersects the graph in at least two distinct points

  $f\left(x\right)=-{\left(x-4\right)}^2-3$

Calculation:

If the slope is found for any two points on increasing interval of function, then the secant line will always move up and to the right because one of the points will always be above and to the right of the other point. Any line moving up and to the right has a positive slope.

d.

To determine

To find: the intervals where $f\left(x\right)$ is decreasing.

Answer to Problem 36E

  $\left(4,\infty \right)$

Explanation of Solution

Given information:

A line is secant to the graph of $f\left(x\right)$ if it intersects the graph in at least two distinct points

  $f\left(x\right)=-{\left(x-4\right)}^2-3$

Calculation:

Plot the graph using a graphing calculator as look for the point of maxima in the graph. The graph of $f\left(x\right)$ is shown below.

  

The graph is decreasing from $x=4$ to $x=\infty$ . So, the interval for which $f\left(x\right)$ decreases is $\left(4,\infty \right)$ .

e.

To determine

To make: a hypothesis about the slope of any secant line that passes through any two points contained in the interval evaluated in part d, and explain your reason.

Answer to Problem 36E

The slope of any line made on an interval where the function is decreasing will be negative.

Explanation of Solution

Given information:

A line is secant to the graph of $f\left(x\right)$ if it intersects the graph in at least two distinct points

  $f\left(x\right)=-{\left(x-4\right)}^2-3$

Calculation:

If the slope is found for any two points on decreasing interval of function, then the secant line will always move down and to the right because one of the points will always be lower and to the right of the other point. Any line moving down and to the right has a negative slope.

The two point selected from the decreasing interval to find the slope are $\left(5,-4\right)$ and $\left(6,-7\right)$ .

Now put the points in the slope formula and evaluate m as follows:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-7-\left(-4\right)}{6-5}\\ =\frac{-7+4}{1}\\ =-3\end{array}$

The slope of the secant line evaluated is $-3$ and thus, the hypothesis made above is proved.