Which function has a greater rate of change over the interval [0,2] when table x=-1,0,1,2 and g(x)=7,5,7,13

answer: =34

Step-by-step explanation:

Example 1: Average rate of change from graph

Let's find the average rate of change of

f

ff over the interval

0

≤

x

≤

9

0≤x≤90, is less than or equal to, x, is less than or equal to, 9:

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

-

2

-

2

-

3

-

3

-

4

-

4

-

5

-

5

1

1

2

2

3

3

4

4

5

5

6

6

7

7

-

2

-

2

-

3

-

3

-

4

-

4

-

5

-

5

-

6

-

6

-

7

-

7

y

y

x

x

y

=

f

(

x

)

y=f(x)

We can see from the graph that

f

(

0

)

=

−

7

f(0)=−7f, left parenthesis, 0, right parenthesis, equals, minus, 7 and

f

(

9

)

=

3

f(9)=3f, left parenthesis, 9, right parenthesis, equals, 3.

Average rate of change

=

f

(

9

)

−

f

(

0

)

9

−

0

=

3

−

(

−

7

)

9

=

10

9

Average rate of change

​

=

9−0

f(9)−f(0)

​

=

9

3−(−7)

​

=

9

10

​

​

Example 2: Average rate of change from equation

Let's find the rate of change of

g

(

x

)

=

x

3

−

9

x

g(x)=x

3

−9xg, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 9, x over the interval

1

≤

x

≤

6

1≤x≤61, is less than or equal to, x, is less than or equal to, 6.

g

(

1

)

=

1

3

−

9

⋅

1

=

−

8

g(1)=1

3

−9⋅1=−8g, left parenthesis, 1, right parenthesis, equals, 1, cubed, minus, 9, dot, 1, equals, minus, 8

g

(

6

)

=

6

3

−

9

⋅

6

=

162

g(6)=6

3

−9⋅6=162g, left parenthesis, 6, right parenthesis, equals, 6, cubed, minus, 9, dot, 6, equals, 162

Average rate of change

=

g

(

6

)

−

g

(

1

)

6

−

1

=

162

−

(

−

8

)

5

=

34

Average rate of change

​

=

6−1

g(6)−g(1)

​

=

5

162−(−8)

​

=34

​

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap