Difference Quotient...3

[nycmathguy]

Chapter 1
Calculus 1

Question 27

Find (3 + h)

f(3 + h) = 4 + 3(3 + h) - (3 + h)^2

f(3 + h) = 4 + 9 + 3h - (9 + 6h + h^2)

f(3 + h) = 4 + 9 + 3h - 9 - 6h - h^2

f(3 + h) = 4 + 3h - 6h - h^2


Find f(3).

f(3) = 4 + 3(3) - (3)^2

f(3) = 4 + 9 - 9

f(3) = 4

We now have:

[4 + 3h - 6h - h^2 - 4]/h

(3h - 6h - h^2)/h

3 - 6 - h

Answer: -3 - h

What does the answer mean?

 

[MathLover1]

correct

What does the answer mean? -> mean you have an expression in terms of h as an answer, and what will be the answer depends on value(s) of h

 

[nycmathguy]

correct

What does the answer mean? -> mean you have an expression in terms of h as an answer, and what will be the answer depends on value(s) of h

Ok. So far so good in terms of calculus.

 

[HallsofIvy]

It means that the average rate of change in f(x) between x and x+h is -3- h. Notice that with regular algebra, the question "what does the instantaneous rate of change", the rate of change with no change at all in x, makes no sense- if x does not change f(x) does not change so there is no "rate of change". But wth the limit concept, we can define the "instantaneous rate of change" as the limit as h goes to 0 and lim -3- h= -3 so the instantaneous rate of change of f(x) at x is -3.

 

[nycmathguy]

It means that the average rate of change in f(x) between x and x+h is -3- h. Notice that with regular algebra, the question "what does the instantaneous rate of change", the rate of change with no change at all in x, makes no sense- if x does not change f(x) does not change so there is no "rate of change". But wth the limit concept, we can define the "instantaneous rate of change" as the limit as h goes to 0 and lim -3- h= -3 so the instantaneous rate of change of f(x) at x is -3.

The answer is just -3 - h. I now know that the answer means the average rate of change.