Definition Let c be a number in the domain D of a function f. Then f (c) is the
• absolute maximum value of f on D if f (c) ≥ f (x) for all x in D.
• absolute minimum value of f on D if f (c) ≤ f (x) for all x in D.
An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of f are called extreme values of f.
Definition The number f (c) is called a
• local maximumvalue of f if c is an interior point and f (c) ≥ f (x) when x is near c, i.e. if there exists a δ > 0 such that f (c) ≥ f (x) for all x ∈ (c − δ, c + δ).
• local minimum value of f if c is an interior point and f (c) ≤ f (x) when x is near c, i.e. if there exists a δ > 0 such that f (c) ≤ f (x) for all x ∈ (c − δ, c + δ).
Note that if we say that something is true near c, we mean that it is true on some open interval containing c.
Extreme Value TheoremIff is continuous ona closed interval[a, b], then there exist x1, x2 ∈ [a, b] such that f (x1) is the absolute maximum value and f (x2) is the absolute minimum value of f on [a, b], i.e.
min
x∈[a,b]
f (x) = f (x1) and max
x∈[a,b]
f (x) = f (x2).
Remarks The proof of the Extreme Value Theorem will be omitted here since it requires the following facts that are beyond the scope of the course:
1. A set S of real numbers is called bounded from above if there is a k ∈ R such that s ≤ k for all s in S.
Here, the element k is called anupper boundof S. A set S of real numbers is calledbounded if there is a m ∈ R such that
|s| ≤ m for all s in S.
A number b is called the supremumof S and it is denoted by b = sup S if (a) b is an upper boundof S, i.e. s ≤ b for all s in S.
(b) if k is another upper bound of S then b ≤ k, i.e. b is the least upper bound of S, so for each ε > 0, there is an s ∈ S such that s > b − ε.
Example Let S = (a, b). Then both sup S = b 6∈ S and inf S = a 6∈ S exist while neither max S nor min S exists.
– A is bounded above by b, and since f is continuous at a, – f is bounded on [a, a], we have a ∈ A and A 6= ∅.
By the completeness of R, there exists c ≤ b such that c = sup A.
Suppose that c < b, by the continuity of f at c, there exists a b − c > δ > 0 such that if x ∈ [a, b] ∩ (x − δ, c + δ), then |f (x) − f (c)| < 1 =⇒ |f (x)| < |f (c)| + 1, i.e. f is bounded by |f (c)| + 1 on [a, b] ∩ (x − δ, c + δ) ⊂ [a, b].
a b
c − δ c c + δ
c +δ 2
Since [a, c +δ
2] ⊂ [a, b] ∩ (x − δ, c + δ), f is bounded on [a, c +δ
2] which implies that c +δ 2 ∈ A, and since c + δ
2 > c, this is a contradiction to that c = sup A.
Hence c = b and f is bounded on [a, b].
4. If f is continuous on [a, b], then there exist x1, x2 ∈ [a, b] so that the supremum and infimum values of f are attained at x2 and x1, respectively, i.e.
inf
x∈[a,b]f (x) = f (x1) and sup
x∈[a,b]
f (x) = f (x2) exist.
Example Let f (x) = x for x ∈ (a, b). Then both sup
x∈(a,b)
f (x) = b and inf
x∈(a,b)f (x) = a exist while neither max
x∈(a,b)f (x) nor min
x∈(a,b)f (x) exists.
Example Let
f (x) =
(x if 0 < x < 1 1/2 if x = 0 or 1 Then both sup
x∈(0,1)
f (x) = 1 and inf
x∈(0,1)f (x) = 0 exist while neither max
x∈(0,1)f (x) nor min
x∈(0,1)f (x) exists.
Fermat’s Theorem If f has a local maximum or minimum at c, and if f0(c) exists, then f0(c) = 0.
Remark Consider the example y = f (x) = x3, for x ∈ R. Note that f0(0) = 0, but f (0) = 0 is neither a maximum nor a minimum value of f. (In other words, the converse of Fermats Theorem is false in general.)
Furthermore, the example y = f (x) = |x| shows that f (0) = 0 is an extreme (minimum) value of f even when f0(0) does not exist.
Rolle’s Theorem Let f be continuous on the closed interval [a, b], differentiable on the open interval (a, b).
Suppose that f (a) = f (b), then there is a number c in (a, b) such that f0(c) = 0, i.e. the tangent line to y = f (x) at (c, f (c)) exists and it is a horizontal line.
Mean Value Theorem Let f be continuous on the closed interval [a, b], differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that f0(c) = f (b) − f (a)
b − a ⇐⇒ f (b) − f (a) = f0(c)(b − a).
Proof: Consider the function
g(x) = f (x) − f (a) − f (b) − f (a)
b − a (x − a).
Since g is continuous on [a, b], differentiable on (a, b) and g(a) = g(b) = 0, so, by the Rolle’s Theorem, there is a c ∈ (a, b) such that
0 = g0(c) = f0(c) − f (b) − f (a)
b − a =⇒ f0(c) = f (b) − f (a) b − a .
Examples
1. Find where the function 3x4− 4x3− 12x2+ 5 is increasing and where it is decreasing.
2. Prove that the equation x3+ x − 1 = 0 has exactly one real solution.
Definition A critical numberof a function f is a number c in the domain of f such that either f0(c) = 0 or f0(c) does not exist.
Remark If f has a local maximum or minimum at c, then c is a critical number of f.
To find an absolute maximum or minimum of a continuous function on a closed interval, we note that either it is local or it occurs at an endpoint of the interval.
Thus the following three-step procedure always works.
The Closed Interval MethodTo find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:
1. Find the values of f at the critical numbers of f in (a, b), the interior of [a, b].
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
The First Derivative Test Suppose that c is a critical number of a continuous function f.
(a) If f0 changes from positive to negative at c, then f has a local maximum at c.
(b) If f0 changes from negative to positive at c, then f has a local minimum at c.
(c) If f0 is positive to the left and right of c, or negative to the left and right of c, then f has no local maximum or minimum at c.
Example Find the local maximum and minimum values of the function g(x) = x + 2 sin x for 0 ≤ x ≤ 2π.
Definition If the graph of f lies above all of its tangents on an interval I, then f is called concave upwardon I. If the graph of f lies below all of its tangents on I, then f is calledconcave downwardon I.
Concavity Test
(a) If f00(x) > 0 on an interval I, then the graph of f is concave upward on I.
Proof For each point a, x ∈ I, consider the function
g(x) = f (x) −f0(a)(x − a) + f (a).
Since g00(x) = f00(x) > 0 and g0(a) = 0,
• g0(x) is increasing on an interval I,
• g0(x) < 0 for x < a,
• g0(x) > 0 for x > a.
This implies that
• g(x) ≥ g(a) = 0 for all x ∈ I and hence
• the graph of f lies above all of its tangents on I, i.e.
the graph of f is concave upward on I.
(b) If f00(x) < 0 on an interval I, then the graph of f is concave downward on I.
Definition A point P on a curve y = f (x) is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P.
The Second Derivative TestSuppose f00 is continuous near c.
(a) If f0(c) = 0 and f00(c) > 0, then f has alocal minimum at c.
(b) If f0(c) = 0 and f00(c) < 0, then f has alocal maximum at c.
Examples
1. Discuss the curve y = f (x) = x4− 4x3 with respect to concavity, points of inflection, and local maxima and minima.
Note that f (x) = x3(x − 4) = 0 when x = 0 or 4; f0(x) = 4x2(x − 3) = 0 when x = 0 or 3;
and f00(x) = 12x(x − 2) = 0 when x = 0 or 2.
(−∞, 0) (0, 2) (2, 3) (3, ∞) f0(x) < 0 < 0 < 0 > 0 f00(x) > 0 < 0 > 0 > 0
f (x)
f00(x) < 0 < 0 < 0 > 0 f (x)
Indeterminate Forms and l’Hospital’s Rule
Definitions 1. If lim
x→af (x) = 0 and lim
x→ag(x) = 0, then the limit
x→alim f (x) g(x)
may or may not exist and it is called anindeterminate form of type 0 0. For example, lim
x→1
x2− x x2− 1, lim
x→0
sin x
x and lim
x→π
sin x
(x − π)2 are indeterminate forms of type 0 0. 2. If lim
x→af (x) = ∞ ( or − ∞) and lim
x→ag(x) = ∞ ( or − ∞), then the limit
x→alim f (x) g(x)
may or may not exist and it is called anindeterminate form of type ∞
∞. For example, lim
x→∞
ln x x − 1, lim
x→∞
x2− 1
2x2+ 1 and lim
x→(π/2)+
tan x
ln (x − π/2) are indeterminate forms of type ∞
∞.
L’Hospital’s Rule Suppose that
• f and g are differentiable and g0(x) 6= 0 on an open interval I that contains a (except possibly at a),
• lim
x→af (x) = 0 and lim
x→ag(x) = 0 (an indeterminate form of type 0 0), or lim
x→af (x) = ±∞ and lim
x→ag(x) = ±∞ (an indeterminate form of type ±∞
∞ ).
If lim
x→a
f0(x)
g0(x) exists (or is ∞ or −∞), then
x→alim f (x)
g(x) = lim
x→a
f0(x) g0(x).
Remark For the special case in which f (a) = g(a) = 0, f0 and g0 are continuous, and g0(a) 6= 0, it is easy to see why l’Hospital’s Rule is true. In fact, using the alternative form of the definition of a derivative, we have
x→alim f0(x)
g0(x) = f0(a)
g0(a) [since f0, g0 are continuous and g0(a) 6= 0]
=
x→alim
f (x) − f (a) x − a
x→alim
g(x) − g(a) x − a
= lim
x→a
f (x) − f (a) g(x) − g(a)
= lim
x→a
f (x)
g(x) [since f (a) = g(a) = 0]
It is more difficult to prove the general version of l’Hospital’s Rule.
Examples Show that 1. lim
x→1
ln x x − 1 = 1 2. lim
x→∞
ex x2 = ∞
3. lim
x→∞
ln x√ x = 0 4. lim
x→0
tan x − x x3 = 1
3
5. lim
x→π+
sin x 1 − cos x = 0 6. lim
x→π+
sin x
1 + cos x = −∞
Other Indeterminate Forms (Types 0 · ∞, ∞ − ∞, 00, ∞0, 1∞)
Examples Show that 1. lim
x→0+x ln x = 0 2. lim
x→1+
1
ln x− 1
x − 1 = 1 2
3. lim
x→0+xx = 1 4. lim
x→∞x1/x= 1
5. lim
x→0+ 1 + sin 4xcot x
= e4
Examples By determining its domain, intercepts, symmetry, asmptotes, intervals of increase of decrease, local maximum or minimum values, concavity and points of inflection (if exists), sketch the graph of the following:
1. y = 2x2
x2− 1 = 2 + 2
x2− 1 2. f (x) = x3
x2+ 1 = x − x x2+ 1
Examples of Optimization Problem
1. A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
Hint: Let x and y be the depth and width of the rectangle (in feet). Then we want to maximize A = xy subject to the constraint: the total length of the fencing is 2400 ft, i.e.
2x + y = 2400 =⇒ A = x(2400 − 2x) for 0 ≤ x ≤ 1200.
20 (or the price function) is p(x) = 350 −10
20(x − 200) = 450 −1
2x and the revenue function is R(x) = xp(x) = 450x −1
2x2 achieves maximum when x = 450.
Definition A function F is called an antiderivative of f on an interval I if F0(x) = f (x) for all x in I.
Theorem If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F (x) + C where C is an arbitrary constant.
Examples Find the most general antiderivative of each of the following functions.
(a) f (x) = sin x (b) f (x) = 1
x
(c) f (x) = xn, n 6= −1 (d) f (x) = 4 sin x + 2x5−√
x x
Examples
1. The following are graphs of a function f and a rough sketch of its antiderivative F satisfying F (0) = 2.
2. A particle moves in a straight line and has acceleration given by a(t) = 6t + 4. Its initial velocity is v(0) = 6 cm/s and its initial displacement is s(0) = 9 cm. Find its position function s(t).
