Calculus
Of One Real Variable – By Pheng Kim Ving |
5.3 |
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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).
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Fig. 1.1
f is increasing on [a, b].
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Fig. 1.2
f is decreasing on [a, b].
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Fig. 1.3
f is non-decreasing on [a, b].
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Fig. 1.4
f is non-increasing on [a, b].
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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.
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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.
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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
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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.
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Fig. 3.1
Chart conveying increasing and decreasing
behavior of f.
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EOS
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.
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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.
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Fig. 4.1
Local maxima occur at x1 and x3.
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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. |
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.
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;
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Fig. 4.2
Discontinuities in this example show
that the continuity in the
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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.
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Fig. 4.3
Local extrema at endpoints.
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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.
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Fig. 5.1
f(0) = 1 is neither a local maximum nor a local
minimum.
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Fig. 5.2 Graph of f(x) = x4 – 2x3 + 1. |
EOS
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.
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Note: The
vertical double line under a value of x in a row |
There are no local maximum or minimum values.
Solution
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