Recall that the slope of a line is positive if, and only if, the line rises from left to right. That is, ifm >0,f(x) =mx+b, and u < v, then
f(v) =mv+b
=mv−mu+mu+b
=m(v−u) +mu+b > mu+b
=f(u). (1.9.1)
We should expect that an analogous statement holds for differentiable functions: iff is differentiable andf0(x)>0 for allxin an interval (a, b), thenf(v)> f(u)
for anyv > uin (a, b). This is in fact the case, although the inference requires establishing a direct connection between slope at a point and the average slope
over an interval, or, in terms of rates of change, between the instantaneous rate of change at a point and the average rate of change over an interval. The mean-value theorem makes this connection.
1.9.1
The mean-value theorem
Recall that the extreme value property tells us that a continuous function on a closed interval must attain both a minimum and a maximum value. Supposef
is continuous on [a, b], differentiable on (a, b), andf attains a maximum value at c with a < c < b. In particular, for any infinitesimaldx, f(c)≥f(c+dx), and so, equivalently,f(c+dx)−f(c)≤0. It follows that ifdx >0,
f(c+dx)−f(c)
dx ≤0, (1.9.2)
and ifdx <0,
f(c+dx)−f(c)
dx ≥0. (1.9.3)
Since both of these values must be infinitesimally close to the same real number, it must be the case that
f(c+dx)−f(c)
dx '0. (1.9.4)
That is, we must havef0(c) = 0. A similar result holds iff has a minimum at
c, and so we have the following basic result.
Theorem 1.9.1. If f is differentiable on (a, b) and attains a maximum, or a minimum, value atc, thenf0(c) = 0.
Now suppose f is continuous on [a, b], differentiable on (a, b), and f(a) =
f(b). If f is a constant function, then f0(c) = 0 for all c in (a, b). If f is not
constant, then there is a pointc in (a, b) at whichf attains either a maximum or a minimum value, and so f0(c) = 0. In either case, we have the following result, known as Rolle’s theorem.
Theorem 1.9.2. Iff is continuous on [a, b], differentiable on (a, b), andf(a) =
f(b), then there is a real number cin (a, b) for which f0(c) = 0.
More generally, supposef is continuous on [a, b] and differentiable on (a, b). Let
g(x) =f(x)−f(b)−f(a)
b−a (x−a)−f(a). (1.9.5)
Note that g(x) is the difference between f(x) and the corresponding y value on the line passing through (a, f(a)) and (b, f(b)). Moroever, g is continuous on [a, b], differentiable on (a, b), and g(a) = 0 =g(b). Hence Rolle’s theorem applies tog, so there must exist a pointcin (a, b) for which g0(c) = 0. Now
g0(c) =f0(x)−f(b)−f(a)
0 1 2 x −2 −1 0 1 2 3 y y=x3−3x+ 1
Figure 1.9.1: Graph off(x) =x3−3x+ 1 with its tangent line atx=
q 4 3 so we must have 0 =g0(c) =f0(c)−f(b)−f(a) b−a . (1.9.7) That is, f0(c) =f(b)−f(a) b−a , (1.9.8)
which is our desired connection between instantaneous and average rates of change, known as themean-value theorem.
Theorem 1.9.3. If f is continuous on [a, b] and differentiable on (a, b), then there exists a real numberc in (a, b) for which
f0(c) = f(b)−f(a))
b−a . (1.9.9)
Example 1.9.1. Consider the functionf(x) =x3−3x+ 1 on the interval [0,2]. By the mean-value theorem, there must exist at least one point c in [0,2] for which
f0(c) =f(2)−f(0) 2−0 =
3−1 2 = 1.
Now f0(x) = 3x2−3, so f0(c) = 1 implies 3c2−3 = 1. Hencec =q4 3. Note that this implies that the tangent line to the graph off at x=q43 is parallel to the line through the endpoints of the graph off, that is, the points (0,1) and (2,3). See Figure1.9.1.
1.9.2
Increasing and decreasing functions
The preceding discussion leads us to the following definition and theorem.
Definition 1.9.1. We say a functionfisincreasingon an intervalIif, whenever
a < b are points in I, f(a) < f(b). Similarly, we say f is decreasing on I if, whenevera < bare points inI,f(a)> f(b).
Now supposef is a defined on an intervalI and f0(x)>0 for every xin I
which is not an endpoint ofI. Then given anyaandbinI, by the mean-value theorem there exists a pointc betweenaand bfor which
f(b)−f(a)
b−a =f
0(c)>0. (1.9.10)
Since b−a >0, this implies that f(b)> f(a). Hencef is increasing onI. A similar argument shows that f is decreasing on I iff0(x)<0 for everyxin I
which is not an endpoint of I.
Theorem 1.9.4. Supposef is defined on an intervalI. If f0(x)>0 for every
xinI which is not an endpoint ofI, thenf is increasing onI. Iff0(x)<0 for everyxin Iwhich is not an endpoint of I, thenf is decreasing onI.
Example 1.9.2. Letf(x) = 2x3−3x2−12x+ 1. Then
f0(x) = 6x2−6x−12 = 6(x2−x−2) = 6(x−2)(x+ 1).
Hence f0(x) = 0 when x = −1 and when x = 2. Now x−2 < 0 for x <2 and x−2 > 0 for x > 2, while x+ 1 < 0 for x < −1 and x+ 1 > 0 when
x > −1. Thus f0(x)>0 when x < −1 and when x > 2, andf0(x)<0 when −1 < x < 2. It follows that f is increasing on the intervals (−∞,−1) and (2,∞), and decreasing on the interval (−1,2).
Note that the theorem requires only that we know the sign off0 at points inside a given interval, not at the endpoints. Hence it actually allows us to make the slightly stronger statement that f is increasing on the intervals (−∞,−1] and [2,∞), and decreasing on the interval [−1,2].
Sincef is increasing on (−∞,−1] and decreasing on [−1,2], the point (−1,8) must be a high point on the graph of f, although not necessarily the highest point on the graph. We say that f has a local maximum of 8 at x = −1. Similarly, f is decreasing on [−1,2] and increasing on [2,∞), and so the point (2,−19) must be a low point on the graph off, although, again, not necessarily the lowest point on the graph. We say that f has a local minimum of−19 at
x= 2. From this information, we can begin to see why the graph off looks as it does in Figure 1.9.2.
Definition 1.9.2. We sayf has alocal maximumat a pointcif there exists an interval (a, b) containingcfor whichf(c)≥f(x) for allxin (a, b). Similarly, we sayf has alocal minimumat a pointcif there exists an interval (a, b) containing
c for whichf(c)≤f(x) for allxin (a, b). We say f has a local extremum atc
−3 −2 −1 0 1 2 3 4 x −40 −30 −20 −10 0 10 20 30 y y= 2x3−3x2−12x+ 1 Figure 1.9.2: Graph off(x) = 2x3−3x2−12x+ 1
We may now rephrase Theorem 1.9.1as follows.
Theorem 1.9.5. Iff is differentiable atcand has a local extremum atc, then
f0(c) = 0.
As illustrated in the preceding example, we may identify local minimums of a function f by locating those points at which f changes from decreasing to increasing, and local maximums by locating those points at which f changes from increasing to decreasing.
Example 1.9.3. Let f(x) =x+ 2 sin(x). Then f0(x) = 1 + 2 cos(x), and so
f0(x)<0 when, and only when,
cos(x)<−1 2. For 0≤x≤2π, this occurs when, and only when,
2π
3 < x < 4π
3 .
Since the cosine function has period 2π, if follows that f0(x) < 0 when, and only when,xis in an interval of the form
2π
3 + 2πn, 4π
3 + 2πn
for n= 0,±1,±2, . . .. Hence f is decreasing on these intervals and increasing on intervals of the form
−2π 3 + 2πn, 2π 3 + 2πn ,
−10 0 10 x −10 0 10 y y=x+ 2 sin(x)
Figure 1.9.3: Graph off(x) =x+ 2 sin(x)
n= 0,±1,±2, . . .. It now follows thatf has a local maximum at every point of the form
x=2π 3 + 2πn and a local minimum at every point of the form
x=4π 3 + 2πn.
From this information, we can begin to see why the graph off looks as it does in Figure1.9.3.
Exercise 1.9.1. Find the intervals where f(x) = x3−6x is increasing and the intervals where f is decreasing. Use this information to identify any local maximums or local minimums off.
Exercise 1.9.2. Find the intervals wheref(x) = 5x3−3x5 is increasing and the intervals where f is decreasing. Use this information to identify any local maximums or local minimums off.
Exercise 1.9.3. Find the intervals wheref(x) =x+ sin(x) is increasing and the intervals where f is decreasing. Use this information to identify any local maximums or local minimumsf.