Question

for a function j(x) the difference quotient is 3x^2 + 3xh + h^2. what is the average rate of change of j(x) over the interval [-1,6]

Answer 1

Average rate of change = 15 + 3h

Explanation:

• Let's say want to find the average rate of change (AROC) over the interval [a, b], where a and b represent numbers.

The average rate of change formula is given by:

`(f(b)-f(a))/(b-a)`

• Essentially, for f(b) and f(a), you substitute the interval values for the function values like 6 for x in f(x) = 2x + 5.

Since we'd plug in 6 for b and -1 for a , we can write the average rate of change in terms of j(x):

`(j(6)-j(-1))/(6-(-1))`

`((3(6)^2+3(6)h+h^2)-(3(-1)^2+3(-1)h+h^2))/(6-(-1))`

To prevent confusion, I can explain what happens at each step:

Step 1:

• Numerator: Square 6 and -1; multiply 3, 6, h; find the product of 3, -1 and h.
• Denominator: Distribute the negative to (-1):

`((3(36)+18h+h^2)-(3(1)+-3h+h^2))/(6+1)`

Step 2:

• Numerator: Multiply 3 and 36; multiply 3 and 1.
• Denominator: Add 6 and 1:

`((108+18h+h^2)-(3-3h+h^2))/(7)`

Step 3:

• Numerator: Distribute the negative to (3 - 3h + h^2):

`((108+18h+h^2)+(-3+3h-h^2))/(7)`

Step 4: Combine like terms on the numerator:

`((108-3)+(18h+3h)+(h^2-h^2))/(7)\\ \\(105+21h)/(7)`

Step 4: Split the equation into two and simplify:

`(105)/(7)+(21h)/(7)\\ \\ 15+3h`

Therefore, 15 + 3h is the average rate of change of j(x) over the interval [-1, 6], given that j(x) = 3x^2 + 3xh + h^2.