|
Calculus
Of One Real Variable – By Pheng Kim Ving |
|
5.3 |
Return
To Contents
Go To Problems & Solutions
|
1. Increasing And Decreasing Functions |
Consider the function f in Fig. 1.1. As x
increases, the graph of f rises,
which means that the value f(x) increases. We
say that f is increasing. We note that
if x1 < x2, then f(x1) < f(x2).
Consider the function f in Fig. 1.2. As x
increases, the graph of f falls,
which means that the value f(x) decreases. We
say that f is decreasing. We note that
if x1 < x2, then f(x1) > f(x2).
|
|
Fig. 1.1
f is increasing on [a, b].
|
|
|
Fig. 1.2
f is decreasing on [a, b].
|
|
|
Fig. 1.3
f is non-decreasing on [a, b].
|
|
|
Fig. 1.4
f is non-increasing on [a, b].
|
Definition 1.1
|
|
Note that for an increasing function f, f(x) changes in
the same direction as x: if x increases then f(x) increases, if x
decreases then f(x)
decreases, and that for a decreasing function f,
f(x) changes in
the opposite direction to x: if x
increases then f(x)
decreases, if x decreases then f(x) increases.
Also note that the notions of increasing and decreasing
behaviors of f on I
don't require the continuity (and thus the differentiability) of f on I.
Go To Problems & Solutions Return To Top Of Page
|
2. Test For Increasing And Decreasing Behaviors |
Let the function y = f(x) be differentiable on an interval I. Suppose that f
'(x) = dy/dx > 0 for all x
in I. Since the rate of
change of f is positive, the change in x and the corresponding change in y must be of the same sign, ie, f must be an
increasing function.
Theorem 2.1
|
|
Proof
We prove part i. The proofs
for all the remaining parts are similar to it. Let x1 and x2 be arbitrary points in I
with x1 < x2.
By the mean-value theorem, there exists c in (x1, x2) such that ( f(x2) – f(x1))/(x2 – x1) = f '(c). Now, c is in I too, so
f '(c) > 0. Also, x2 – x1 > 0. Consequently,
f(x2) – f(x1) > 0, ie, f(x1) < f(x2). Thus f is increasing on I.
EOP
Go To Problems & Solutions Return To Top Of Page
|
3. Finding Intervals Of Increase And Decrease |
An interval of increase of f is one where f is increasing. An interval of decrease of f is one where f is decreasing.
Example 3.1
Find the intervals of increase and decrease of the function f(x) = x3 – 3x2 + 5.
Solution
|
|
Fig. 3.1
Chart conveying increasing and decreasing
behavior of f.
|
EOS
Chart Conveying The Increasing And Decreasing Behaviors
Remark that on an interval, f '(x) > 0 if
the number of “ – ” signs in that
interval above the row for f '(x) is 0 or even, and
f '(x) < 0 if that number is odd.
Go To Problems & Solutions Return To Top Of Page
|
4. Test For Local Extrema |
As we saw in Section
5.2 Remark 4.2, not every critical point yields a local extremum. We're now
going to discuss a test
for determining if a critical point yields a local maximum, or a local minimum,
or neither. Recall that local extrema can
occur only at endpoints and critical points.
Refer to Fig. 4.1. We have a local
maximum of f at x1. We observe
that f is increasing on the left of x1 and decreasing on
the right; thus f '(x)
> 0 for all points x near and to
the left of x1 and f
'(x) < 0 for all points x near and to the right of x1.
We also have a local maximum at x3. Both x1 and x3 are critical
points: f '(x1) = 0, f
'(x3) doesn't
exist.
|
|
Fig. 4.1
Local maxima occur at x1 and x3.
|
Theorem 4.1 – The First-Derivative
Test
|
Suppose that x0 is a
critical point of the function f and that f is differentiable on some open interval
containing x0 except i. If f '(x) > 0 on some interval (x1, x0) and f '(x) < 0 on some interval (x0, x2), then f has a local maximum at x0. ii. If f '(x) < 0 on some interval (x1, x0) and f '(x) > 0 on some interval (x0, x2), then f has a local minimum at x0. |
Proof
EOP
The test is called a first-derivative
test because it involves the first
derivative f
' of a function f.
Note that it's a test for
local extrema.
Remarks 4.1
i. f must be
differentiable at all nearby points of x0
(this is needed by the conditions in parts i and ii of the theorem). It
may or may not be differentiable at x0. See the points x1, x2, x3, and x4 in Fig. 4.1.
ii. f must be
continuous at x0. For example, in Fig. 4.2, f is discontinuous at x1, x2, and x3. For each of
these points, f is
increasing
on some interval to the left of it and decreasing on some interval to the right.
Now, f(x1) is a local
maximum;
|
|
Fig. 4.2
Discontinuities in this example show
that the continuity in the
|
f(x2) isn't, because it doesn't exist; and f(x3) isn't, because,
even though f(x) < f(x3) for all x to the near right of x3,
f(x) > f(x3) for all x to the near
left. So, if f is discontinuous at x0 (and the rest of the hypothesis of the theorem is
satisfied), f(x0) may or may not be a local extremum.
iii. If either f
'(x)
> 0 or f '(x) < 0 on both the near
left and the near right of x0, f has no local extremum at x0. For
example, see the point x5 in Fig. 4.1.
Test On Endpoints For Local Extrema
See Fig. 4.3. Suppose a is the left endpoint of dom( f ) and f
is continuous at a and differentiable on the
right of a for
some distance. If f '(x) > 0, respectively f '(x) < 0, on some interval (a,
x2) (like the case for f1, respectively f2, in Fig.
4.3), then f has a local minimum, respectively a
local maximum, at a. Suppose b is the right endpoint of dom( f ) and f
is continuous at b and differentiable on the
left of b for some distance. If f '(x) > 0, respectively f
'(x)
< 0, on some
interval (x1, b) (like the
case for f1, respectively f2), then f has a local maximum, respectively a local
minimum, at b. The
proof for this result is similar to that of theorem 4.1.
|
|
Fig. 4.3
Local extrema at endpoints.
|
Go To Problems & Solutions Return To Top Of Page
|
5. Locating And Classifying Local Extrema |
Locating local extrema means
locating or finding critical points which each yields a local extremum.
Classifying a local
extremum means classifying or determining it as either a local maximum or a
local minimum.
Example 5.1
Solution
The graph of f is sketched in Fig. 5.2.
|
|
Fig. 5.1
f(0) = 1 is neither a local maximum nor a local
minimum.
|
|
|
Fig. 5.2 Graph of f(x) = x4 – 2x3 + 1. |
EOS
Remark 5.1
The chart in Fig. 5.1 uses the test
for increasing and decreasing behaviors to determine the intervals of increase
and
decrease of f. From the chart we determine
whether a critical point yields a local extremum or not. The first-derivative
test is implicitly used to confirm that f(3/2) =
–11/16 is a local minimum.
|
Problems & Solutions |
1. Find the intervals of increase and decrease of the function f(x) = x2 + 2x + 2.
Solution
2. Determine the intervals in which the function y = x4 – 8x2 + 16 is increasing and decreasing.
Solution
3. Find the intervals of increase and decrease of y = 1/(1 + x2). Find the limits at infinity of y. Sketch the graph of y.
Solution
We have y' = –2x/(1
+ x2)2. So y' = 0 at x
= 0. We have y(0) = 1/(1 + 02) = 1.
Since the denominator of y' is always
positive, the sign of y' is the same as that of its
numerator –2x. The intervals of increase and
decrease of y are shown in
the chart below.
The limits at infinity of y are:
The graph of y is sketched below.
4. For each of the following functions:
- Find limits at infinity and
infinite limits,
- Use the first-derivative test to locate
critical points,
- Use a test and a chart to
determine the intervals of increase and decrease,
- Classify local extreme values if
any, and determine if any of them are absolute extreme values.
Then sketch the graph of the
function.
Solution
so the critical points are x = –1 and x = 1. We have f(–1)
= (–1)/(1 + (–1)2)
= –1/2 and f(1) = 1/(1 + 12) = 1/2. The
sign of f
'(x)
is the same as that of the product (1 + x)(1 – x).
Local maximum value: f(1) = 1/2, local minimum value: f(–1) = –1/2;
absolute maximum value: f(1) = 1/2,
absolute minimum value: f(–1)
= –1/2.
|
|
Note: The
vertical double line under a value of x in a row |
There are no local maximum or minimum values.
Solution
Return To Top Of Page Return To Contents