Download Average Rate of Change and the Difference Quotient

Average Rate of Change and the Di¤erence Quotient
Math 131
Summer 2011
In Section 1.3, p. 16, the average rate of change of a function was introduced. Speci…cally, if y = f (x)
is a function de…ned on an interval [a; b], then the average rate of change of the function on the interval [a; b]
is given by the value of the quotient
f (b) ¡ f (a)
.
b¡ a
Notice that the quotient above can be viewed as the slope
p of the straight line that connects the points (a; f(a))
and (b; f (b)). For example, consider the graph of y = x given below (in red). Let a = 1 and b = 9. Then
(a; f (a)) = (1; 1) and (b; f (b)) = (9; 3).
p
The average rate of change of the function f (x) = x on the interval [1; 9] is given by
p
p
9¡ 1
3¡1
2
1
=
= = .
9¡ 1
9¡1
8
4
p
The green straight line has equation y = 14 x + 34 and intersects the function
f (x) = x at the points (1; 1)
p
and (9; 3). Clearly the average rate of change of the function f (x) = x (which is 14 ) is the slope of the line
y = 14 x + 34 . (Recall that if a straight line is given in slope-intercept form y = mx + b, then the slope is m.
Therefore, the slope of y = 14 x + 34 is 14 .)
f (b) ¡f (a)
The quotient
is very important and is called the di¤erence quotient of the function y = f (x)
b¡a
on the interval [a; b]. We will use a certain variation of this form that is more common in calculus. A
derivation of this form is given below.
Let us …x a point x = a on the x¡axis. Pick an h > 0. (So h is a positive number.) What if we
wanted to compute the average rate of change (that is, the di¤erence quotient) of the function y = f (x) on
the interval [a; a + h]? We would get
f (a + h) ¡ f (a)
f (a + h) ¡ f (a)
=
.
(a + h) ¡ a
h
1
Now suppose h < 0. (So h is a negative number.) Now let us compute the average rate of change of the
function y = f (x) on the interval [a + h; a]. (Notice that a + h < a since h is a negative number.) We
would get
f (a) ¡ f (a + h)
f(a) ¡ f (a + h)
¡(f (a + h) ¡ f (a))
f (a + h) ¡ f (a)
=
=
=
a ¡ (a + h)
¡h
¡h
h
which has the same form as the di¤erence quotient that we derived before. So we conclude that if h > 0 or
if h < 0, the average rate of change of the function y = f (x) on the interval [a; a + h] or [a + h; a] is given
by the quotient
f (a + h) ¡ f (a)
.
h
This is the form of the di¤erence quotient that is the nicest for calculus! Normally, the value a is simply
replaced by x. If we do this replacement, the di¤erence quotient would have the form
f (x + h) ¡ f (x)
h
which would represent the average value of the function y = f (x) on the interval [x; x + h] or [x + h; x] if
h > 0 or h < 0 respectively.
It is important for students to be able to compute and simplify the di¤erence quotient for various types
of functions. The purpose of the remainder of this handout is to provide a couple of examples of how to
compute and simplify the di¤erence quotient. These problems will test your algebra skills, but there is no
calculus here, just algebra. You will need to do these types of problems on Worksheet #2.
Example 1 Compute and simplify the di¤erence quotient for the function f (x) = 5x + 7.
f (x+h)¡ f (x)
To do this, we simply form
for our given function f (x) and then simplify as much as possible.
h
Remember that the expression f (x + h) means that you are evaluating the function f (x) at the expression
x + h. Wherever you see an ‘x ’ in the formula for f (x), replace it with an ‘x + h’. We have
f (x + h) ¡ f (x)
h
5(x + h) + 7 ¡ (5x + 7)
h
5x + 5h + 7 ¡ 5x ¡ 7
=
h
5h
=
h
= 5
(cancel the h since h 6= 0
=
We cannot simplify further, and so we have that
f (x + h) ¡ f (x)
= 5.
h
Note that this says that the average rate of change of the function f (x) = 5x + 7 is exactly 5 on both the
intervals [x; x + h] or [x + h; x] if h > 0 or h < 0 respectively.
Example 2 Compute and simplify the di¤erence quotient for the function f (x) = 3x 2 ¡ 2.
f (x+h)¡ f (x)
As in the previous example, we simply form
for our given function f (x) and then simplify
h
as much as possible. As before, remember that the expression f (x + h) means that you are evaluating the
function f (x) at the expression x + h. Wherever you see an ‘x ’ in the formula for f (x), replace it with an
‘x + h’. We have
f (x + h) ¡ f (x)
h
=
=
3(x + h)2 ¡ 2 ¡ (3x 2 ¡ 2)
h
3(x 2 + 2xh + h 2 ) ¡ 2 ¡ 3x2 + 2
h
2
=
=
=
=
We cannot simplify further, and so
3x 2 + 6xh + 3h2 ¡ 2 ¡ 3x2 + 2
h
6xh + 3h 2
h
h(6x + 3h)
h
6x + 3h
(cancel the h since h 6= 0)
f (x + h) ¡ f (x)
= 6x + 3h.
h
The expression 6x + 3h is the …nal answer. You do not have to do anything further than this. Note that
this says that the average rate of change of the function f(x) = 3x 2 ¡ 2 is exactly 6x + 3h on intervals
[x; x + h] or [x + h; x] if h > 0 or h < 0 respectively. For example, if x = ¡2 and h = 5, then the average
rate of change of the function f (x) = 3x 2 ¡ 2 on the interval [¡2; 3] is 6 ¢ (¡2) + 3 ¢ 5 = ¡12 + 15 = 3. What
would be the average rate of change on the interval [0; 3]? (Hint: x = 0 and h = 3 or, alternately, x = 3
and h = ¡3.)
p
Example 3 Compute and simplify the di¤erence quotient for the function f (x) = 3x ¡ 2.
f (x+h) ¡f (x)
Again, we simply form
for our given function f (x) and then simplify as much as possible.
h
In the derivation below, notice that on line 3 we are multiplying both numerator and denominator by the
conjugate to get rid of the square roots in the numerator. (This process is called rationalizing the numerator.)
We have
p
p
f (x + h) ¡ f (x)
3(x + h) ¡ 2 ¡ 3x ¡ 2
=
h
h
p
p
3x + 3h ¡ 2 ¡ 3x ¡ 2
=
h
p
p
µp
¶ µp
¶
3x + 3h ¡ 2 ¡ 3x ¡ 2
3x + 3h ¡ 2 + 3x ¡ 2
=
¢ p
p
h
3x + 3h ¡ 2 + 3x ¡ 2
3x + 3h ¡ 2 ¡ (3x ¡ 2)
=
¡p
p
¢
h
3x + 3h ¡ 2 + 3x ¡ 2
3x + 3h ¡ 2 ¡ 3x + 2
¡p
¢
=
p
h
3x + 3h ¡ 2 + 3x ¡ 2
3h
=
¡p
p
¢
h
3x + 3h ¡ 2 + 3x ¡ 2
3
= p
p
(cancel the h since h 6= 0)
3x + 3h ¡ 2 + 3x ¡ 2
We cannot simplify further, and so
The expression
f (x + h) ¡ f (x)
3
= p
p
.
h
3x + 3h ¡ 2 + 3x ¡ 2
p
3 p
3x+3h ¡2+ 3x¡ 2
is the …nal answer.
Example 4 Compute and simplify the di¤erence quotient for the function f (x) =
Form
f (x+h)¡ f (x)
h
x
.
x+1
for our given function f (x) and then simplify as much as possible. We have
f (x + h) ¡ f (x)
h
=
=
=
x+ h
x+h+1
¡
x
x+1
(l.c.d. = (x + h + 1)(x + 1))
h
( x+h)(x+1)
(x+h+1)(x+1)
¡
x(x+h +1)
(x+h +1)(x+ 1)
h
(x+h)( x+1)¡ x(x+h+1)
(x+ h+1)( x+1)
h
3
(subtract numerator fractions)
=
=
=
=
=
The expression
(x + h)(x + 1) ¡ x(x + h + 1)
h(x + h + 1)(x + 1)
(simplify the complex fraction)
x2 + x + hx + h ¡ (x2 + hx + x)
h(x + h + 1)(x + 1)
2
x + x + hx + h ¡ x2 ¡ hx ¡ x
h(x + h + 1)(x + 1)
h
h(x + h + 1)(x + 1)
1
(cancel the h since h 6= 0)
(x + h + 1)(x + 1)
1
(x+h+ 1)(x+1)
is the …nal answer since we cannot simplify further.
Remark 5 How do you know that you are done simplifying the di¤erence quotient? This question is the
source of a lot of anxiety for many students. As a general rule, if you can plug in h = 0 into your answer and
not get an expression that requires you to divide by 0, then you know that you have simpli…ed the di¤erence
quotient su¢ciently. For example, consider the answer we obtained in the last example. If we plug in h = 0
into the …nal answer we get
1
1
=
.
(x + 0 + 1)(x + 1)
(x + 1)2
There is no division by 0. Note however that plugging in h = 0 in any previous step will result in an
expression that requires you to divide by 0!
4