Calculate and interpret rate of change over a specified interval - 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

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

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

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

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