Average Rate of Change: Calculus Homework Help

chops369
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Homework Statement


I have just started my first ever calculus course, and I'm having a little trouble with a simple rate of change problem.

It says: Find the average rate of change of the given function between the following pairs of x-values.

The given function is f(x) = x2+ x
The given values are x=1 and x=3

Homework Equations


Aren't I supposed to make use of the equation f(x+h) - f(x) / h ?
I don't really understand what h is supposed to be.

The Attempt at a Solution


I checked what the answer should be and it shows: f(3) - f(1) / 2 = 12-2 / 5 = 5

I understand the algebra and how it equates to the answer 5, but where did the two in the denominator come from? I feel like I'm not understanding some fundamental aspect of this problem and rates of change in general.
 
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To find the average rate of change between point (a, f(a)) and (b, f(b)), you use
(f(b) - f(a))/(b - a). This is basically the same as finding the slope between two points m = (y2 - y1)/(x2 - x1), which you should be familiar with.

There's really no point in using the difference quotient (f(x+h) - f(x))/h, which is an expression, not an equation, for this problem.
 
Bohrok said:
To find the average rate of change between point (a, f(a)) and (b, f(b)), you use
(f(b) - f(a))/(b - a). This is basically the same as finding the slope between two points m = (y2 - y1)/(x2 - x1), which you should be familiar with.

There's really no point in using the difference quotient (f(x+h) - f(x))/h, which is an expression, not an equation, for this problem.

Oh, ok. That makes sense. For some reason the equation you mentioned I should use isn't shown in my book.
 
chops369 said:
Oh, ok. That makes sense. For some reason the equation you mentioned I should use isn't shown in my book.
The expression Bohrok mentioned, namely (f(b) - f(a))/(b - a). An equation has an = sign between two expressions.
 
And, taking h= b-a, the difference between the two points, b= a+ h so the formula you cite, (f(a+h)- f(a))/(h)f(b)- f(a)/(b- a), becomes (f(b)- f(a))/(b- a).
 
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