Rate of change of a function

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Graph 1

Above is the graph of a function. The average rate of change of over the interval is $-\frac{4}{5}$ . Which of these values comes closest to being a possible value of ?

Explanation

The average rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Restated, it is the slope of the line that passes through and .

To find the correct value of that answers this question, it suffices to examine the line with slope $-\frac{4}{5}$ through and find the point among those given that is closest to the line. This line falls 4 units for every 5 horizontal units, so the line looks like this:

Graph 1

The -coordinate of the point of intersection is closer to 2 than to any other of the values in the other four choices. This makes 2 the correct choice.

Graph 1

Above is the graph of a function. The average rate of change of over the interval is $-\frac{4}{5}$ . Which of these values comes closest to being a possible value of ?

Explanation

The average rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Restated, it is the slope of the line that passes through and .

Graph 1

The -coordinate of the point of intersection is closer to 2 than to any other of the values in the other four choices. This makes 2 the correct choice.

Graph 1

Above is the graph of a function , which is defined and continuous on $(-\infty, \infty)$ . The average rate of change of on the interval is 4. Estimate .

Explanation

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Examine the figure below:

Graph 1

The graph passes through the point , so . Therefore,

$4 = \frac{f(5)-f(1)}{5-1}$

and, substituting,

$4 = \frac{f(5)-2}{4}$

Solve for using algebra:

$\frac{f(5)-2}{4} \cdot 4 = 4 \cdot 4$

the correct response.

Define $f(x) = x^{2}+ x$ .

Give the average rate of change of over the interval .

Explanation

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(3)-f(2)}{3-2} = \frac{f(3)-f(2)}{1} = f(3)-f(2)$

Evaluate and by substitution:

$f(x) = x^{2}+ x$

$f(3) = 3^{2}+ 3= 9+ 3 = 12$

$f(2) = 2^{2}+ 2 = 4+ 2 = 6$

$\frac{f(3)-f(2)}{3-2} = 12 - 6 = 6$ ,

the correct response.

Graph 1

Above is the graph of a function , which is defined and continuous on $(-\infty, \infty)$ . The average rate of change of on the interval is 4. Estimate .

Explanation

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Examine the figure below:

Graph 1

The graph passes through the point , so . Therefore,

$4 = \frac{f(5)-f(1)}{5-1}$

and, substituting,

$4 = \frac{f(5)-2}{4}$

Solve for using algebra:

$\frac{f(5)-2}{4} \cdot 4 = 4 \cdot 4$

the correct response.

Define $f(x) = x^{2}+ x$ .

Give the average rate of change of over the interval .

Explanation

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(3)-f(2)}{3-2} = \frac{f(3)-f(2)}{1} = f(3)-f(2)$

Evaluate and by substitution:

$f(x) = x^{2}+ x$

$f(3) = 3^{2}+ 3= 9+ 3 = 12$

$f(2) = 2^{2}+ 2 = 4+ 2 = 6$

$\frac{f(3)-f(2)}{3-2} = 12 - 6 = 6$ ,

the correct response.

Graph 1

Above is the graph of a function . Estimate the rate of change of on the interval

Explanation

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Refer to the graph of the function below:

Graph 1

The graph passes through and .

. Thus,

$r = \frac{f(-1.5)-f(-2)}{-1.5 - (-2)} = \frac{5-2}{0.5} = \frac{3}{0.5} = 6$ ,

the correct response.

Graph 1

Above is the graph of a function . Estimate the rate of change of on the interval

Explanation

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Refer to the graph of the function below:

Graph 1

The graph passes through and .

. Thus,

$r = \frac{f(-1.5)-f(-2)}{-1.5 - (-2)} = \frac{5-2}{0.5} = \frac{3}{0.5} = 6$ ,

the correct response.

Define $f(x) = \left\{\begin{matrix} x^{2}+x ,&x \le 0 \ x^{2}-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

$\frac{1}{2}$

$-\frac{1}{2}$

Explanation

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(1)-f(-1)}{1-(-1)} = \frac{f(1)-f(-1)}{2}$

Evaluate using the definition of for :

$f(x) = x^{2}-x$

$f(1) = 1^{2}-1 = 1 - 1 = 0$

Evaluate using the definition of for $x\le 0$ :

$f(x) = x^{2}+x$

$f(-1) = (-)1^{2}+ (-1) = 1 - 1 = 0$

The average rate of change is therefore

$\frac{f(1)-f(-1)}{2} = \frac{0-0}{2} = 0$ .

Define $f(x) = \left\{\begin{matrix} x^{2}+x ,&x \le 0 \ x^{2}-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

$\frac{1}{2}$

$-\frac{1}{2}$

Explanation

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(1)-f(-1)}{1-(-1)} = \frac{f(1)-f(-1)}{2}$

Evaluate using the definition of for :

$f(x) = x^{2}-x$

$f(1) = 1^{2}-1 = 1 - 1 = 0$

Evaluate using the definition of for $x\le 0$ :

$f(x) = x^{2}+x$

$f(-1) = (-)1^{2}+ (-1) = 1 - 1 = 0$

The average rate of change is therefore

$\frac{f(1)-f(-1)}{2} = \frac{0-0}{2} = 0$ .

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