Rate of change of a function - HiSET

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Question

What is the slope of the line given by the following table?

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

Answer

Given two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

the formula for a slope is

$m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$ .

Thus, since our given table is

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

$m = \frac{3-2}{1-0}=\frac{1}{1} = 1$ .

Hence, the slope of the line generated by the table is

.

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Question

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

Give the average rate of change of over the interval .

Answer

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.

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Question

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 .

Answer

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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Question

The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points.

Answer

The rate of change between two points on a curve can be approximated by calculating the change between two points.

Let be the coordinates of the first point and be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let be the first point and be the second point. Substitute in the values from these coordinates:

$\frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

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Question

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 ?

Answer

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:

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.

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Question

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 .

Answer

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:

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.

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Question

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

Answer

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:

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.

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Question

What is the slope of the line given by the following table?

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

Answer

Given two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

the formula for a slope is

$m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$ .

Thus, since our given table is

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

$m = \frac{3-2}{1-0}=\frac{1}{1} = 1$ .

Hence, the slope of the line generated by the table is

.

Compare your answer with the correct one above

Question

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

Give the average rate of change of over the interval .

Answer

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.

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Question

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 .

Answer

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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Question

The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points.

Answer

The rate of change between two points on a curve can be approximated by calculating the change between two points.

Let be the coordinates of the first point and be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let be the first point and be the second point. Substitute in the values from these coordinates:

$\frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

Compare your answer with the correct one above

Question

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 ?

Answer

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:

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.

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Question

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 .

Answer

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:

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.

Compare your answer with the correct one above

Question

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

Answer

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:

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.

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Question

What is the slope of the line given by the following table?

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

Answer

Given two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

the formula for a slope is

$m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$ .

Thus, since our given table is

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

$m = \frac{3-2}{1-0}=\frac{1}{1} = 1$ .

Hence, the slope of the line generated by the table is

.

Compare your answer with the correct one above

Question

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

Give the average rate of change of over the interval .

Answer

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.

Compare your answer with the correct one above

Question

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 .

Answer

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

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

Compare your answer with the correct one above

Question

The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points.

Answer

The rate of change between two points on a curve can be approximated by calculating the change between two points.

Let be the coordinates of the first point and be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let be the first point and be the second point. Substitute in the values from these coordinates:

$\frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

Compare your answer with the correct one above

Question

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 ?

Answer

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:

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.

Compare your answer with the correct one above

Question

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 .

Answer

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:

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.

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