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

​

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation:

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