Expert: Paul Klarreich - 1/14/2006

Question
There are 3 questions that have 3 parts to them.
The average rate of change between points given.  I've managed to answer the first section.  I'm having difficulties with the 2nd and 3rd sections.  The points are:
(ii) a, f(a) and b, f(b)
(iii) x, f(x) and x+h, f(x+h)

The problems are:
f(x)=5x-4
f(x)= 1/2x+5/2 and
f(x)= xsq. +1

I've tried several different ways and am just frustrating myself.

Answer
Hi, Rebecca,

Re: finding the average rage of change of two points

Yes, I get pretty angry, too, when points change.

HOWEVER, regarding your:

Question:  There are 3 questions that have 3 parts to them.
The average rate of change between points given. I've managed to answer the first section. I'm having difficulties with the 2nd and 3rd sections. The points are:
(ii) a, f(a) and b, f(b)
(iii) x, f(x) and x+h, f(x+h)

The problems are:
f(x)=5x-4
f(x)= 1/2x+5/2 and
f(x)= xsq. +1
----------------------------

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

I think your question is asking for the slope of the line passing through the points.  So you really want your old friend, the slope formula:
y2 - y1
-------
x2 - x1
Or, if the points are (a,f(a)) and (b,f(b))-- your part (ii)

f(b) - f(a)
-----------
  b - a

OR to use the format you will usually want in calculus:(part iii)

f(x + h) - f(x)
---------------
      h

All of these, naturally, represent different form of the same thing.   So, if you want to apply the first of these:

                     f(b) - f(a)
AVG rate of change =  -----------
                       b - a

You simply write  f(b), f(a) and put them into this expression, then you simplify.  (Mostly you are just practicing the function notation, which requires good parenthesizing more than anything else.)  That last sentence didn't have to be parenthesized, of course.

If f(x)=5x-4   then:

f(a) = 5a - 4
f(b) = 5b - 4

f(b) - f(a)   5b - 4 - (5a - 4)   5b - 4 - 5a + 4
----------- = ----------------- = --------------- =
  b - a           b - a               b - a
5b - 5a     5(b - a)
--------- = -------- = 5
 b - a      (b - a)


If f(x)= 1/2x+5/2,  then (did you write this carefully?)

f(b) - f(a)   1/2b + 5/2 - (1/2a + 5/2)
----------- = ------------------------- =
  b - a              b - a
1/2b + 5/2 - 1/2a - 5/2    1/2b - 1/2a     a - b      -1
------------------------ = ----------- = ---------- = ---
       b - a                b - a       2ab(b - a)   2ab


And if f(x)= x^2 + 1  (that's how to write 'x-squared')

f(b) - f(a)   b^2 - a^2   (b - a)(b + a)
----------- = --------- = -------------- = b + a  or a + b
  b - a        b - a         b - a

Seems you just do the same stuff over and over, right?

Using the third format: notice that  x+h - x = h

f(x + h) - f(x)
---------------
      h

If f(x)=5x-4   then:

f(x + h) = 5(x + h) - 4  and we don't have to rewrite f(x)

f(x + h) - f(x)   5(x + h) - 4 - (5x - 4)
--------------- = ----------------------- =
      h                    h
5x + 5h - 4 - 5x + 4   5h
-------------------- = -- = 5, same as before
       h               h

If f(x)= 1/2x+5/2,  then (Sure you wrote this carefully?)

f(x + h) = 1/(2(x + h))  and we don't have to rewrite f(x)

f(x + h) - f(x)   1/(2(x + h)) + 5/2 - (1/2x + 5/2)
--------------- = --------------------------------- =
      h                        h

1/(2(x + h)) - 1/2x
------------------- = (using an LCD of 2x(x + h))
     h
x - (x + h)     - h           - 1
----------- = ---------- = ---------
2xh(x + h)    2xh(x + h)   2x(x + h)

And if f(x)= x^2 + 1

f(x + h) = (x + h)^2 + 1 which = x^2 + 2xh + h^2 + 1

f(x + h) - f(x)   x^2 + 2xh + h^2 + 1 - (x^2 + 1)
--------------- = ------------------------------- =
      h                       h
x^2 + 2xh + h^2 + 1 - x^2 - 1   2xh + h^2
----------------------------- = ---------- =
           h                       h

h(2x + h)
---------- = 2x + h
  h

Now some advice for the near future.  You will be using these forms in applying the 'Definition of the Derivative.'  At the end, you will be doing one of these things:

Letting  b -> a  (I.E. letting b become a.)  In that case, b-a would be zero and you must not have it in the denominator of a fraction, so you must get it canceled using proper algebra.  If it doesn't get canceled, your algebra is no good.

Letting  h -> 0  (I.E. letting h become 0.)  In that case, h would be zero and you must not have it in the denominator of a fraction, so you must get it canceled using proper algebra.  If it doesn't get canceled, your algebra is no good.

Yes, these sound the same because they really are the same, as you will discover.  Happy differentiating!

Paul