Definitions Let c be a number in the domain D of a function f. Then f (c) is the (a) absolute maximum value of f on D, i.e. f (c) = max
x∈D f (x), if f (c) ≥ f (x) for all x in D.
(b) absolute minimum value of f on D, i.e. f (c) = min
x∈Df (x), if f (c) ≤ f (x) for all x in D.
(c) absolute extremum value of f, if f (c) is either an absolute maximum of minimum value of f on D.
(d) local maximum value of f if f (c) ≥ f (x) when x ∈ D is near c, i.e. there exists δ > 0 such that f (c) ≥ f (x) for all x ∈ D ∩ (c − δ, c + δ).
(e) local minimum value of f if f (c) ≤ f (x) when x ∈ D is near c, i.e. there exists δ > 0 such that f (c) ≤ f (x) for all x ∈ D ∩ (c − δ, c + δ).
(f) local extremumvalue of f if f (c) is either a local maximum or minimum value of f.
(g) A graph is said to be concave up at a point if the tangent line to the graph at that point lies below the graph in the vicinity of the point andconcave down at a point if the tangent line lies above the graph in the vicinity of the point.
(h) A point where the concavity changes (from up to down or down to up) is called a point of inflection.
Definitions Let c be a number in the domain D of a function f. Then c is called a critical number of f if either f0(c) = 0 or f0(c) does not exist.
Extreme Value Theorem Let f be continuous on [a, b]. Then there exist x1, x2 ∈ [a, b] such that
min
[a,b] f = f (x1) ≤ f (x) ≤ f (x2) = max
[a,b] f ∀ x ∈ [a, b].
Fermat’s Theorem Let f be continuous on [a, b]. If f has a local extremum at c ∈ (a, b), and if f0(c) exists, then f0(c) = 0,
i.e. If f has a local extremum at an interior point c ∈ (a, b), then either f is not differentiable at c, or f is differentiable at c, and f0(c) = 0.
Proof Suppose that
lim
h→0
f (c + h) − f (c)
h = f0(c) > 0,
since = f0(c)
2 > 0, there exists δ > 0 such that if 0 < |h| < δ then
f (c + h) − f (c)
h − f0(c)
< f0(c)
2 ⇔ f0(c)
2 > f (c + h) − f (c)
h − f0(c) > −f0(c) 2
=⇒ f (c + h) − f (c)
h > f0(c) 2 > 0
=⇒
(f (c + h) − f (c) > 0 for all δ > h > 0 f (c + h) − f (c) < 0 for all 0 > h > −δ
⇔
(f (c + h) − f (c) > 0 for all δ > h > 0 f (c − h) − f (c) < 0 for all δ > h > 0
=⇒ f (c − h) < f (c) < f (c + h) for all δ > h > 0
=⇒ f increases through c.
Thus f (c) is not a local extremum of f.
Similarly, if f0(c) < 0 then f decreases through c and f (c) is not a local extremum of f.
Hence, if f has a local extremum at an interior point c ∈ (a, b) thenf0(c) = 0.
Remark The local extremum of a continuous function f occurs only at a critical number of f.
Remark To find an absolute extremum of a continuous function f on a closed interval I = [a, b], we note that either it is local or it occurs at an endpoint of the interval I.
Rolle’s Theorem Let f be continuous on I = [a, b] and let f be differentiable on (a, b). Suppose that f (a) = f (b). Then there is a number c ∈ (a, b) such that f0(c) = 0.
Proof
Case 1 Suppose that f (x) = f (a) = f (b) for all x ∈ [a, b] =⇒ f0(x) = 0 for all x ∈ (a, b).
Case 2 Suppose there exits x ∈ (a, b) such that f (x) 6= f (a) = f (b) =⇒ there exists c ∈ (a, b) such that f (c) is an extremum value of f. Therefore, f0(c) = 0 by the Fermat’s Theorem.
The Mean Value Theorem Let f be continuous on I = [a, b] and let f be differentiable on (a, b). Then there is a number c ∈ (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 defined by
g(x) = f (x) − f (a) − f (b) − f (a)
b − a (x − a) if x ∈ [a, b].
Since g is continuous on I = [a, b], differentiable on (a, b) and satisfies that g(a) = g(b) = 0, there exists c ∈ (a, b), by the Rolle’s Theorem, such that
0 = g0(c) = f0(c) − f (b) − f (a)
b − a =⇒ f0(c) = f (b) − f (a) b − a .
Remarks Let f be differentiable on (a, b). Suppose that f0(x) 6= 0 for all x ∈ (a, b). Then (a) f is 1 − 1 on (a, b).
Proof For any x1 6= x2 ∈ (a, b), there exits c lying between x1 and x2, by the Mean Value Theorem, such that
f (x1) − f (x2) = f0(c)(x1− x2) 6= 0 =⇒f (x1) 6= f (x2) =⇒ f is 1 − 1 on (a, b).
(b) If f0(x) > 0 for all x ∈ (a, b), then f is increasing on (a, b), i.e. f (x1) < f (x2) for any x1 < x2 ∈ (a, b).
Proof For any x1 < x2 ∈ (a, b), there exits c ∈ (x1, x2), by the Mean Value Theorem, such that
f (x1) − f (x2) = f0(c)(x1− x2) < 0 =⇒f (x1) < f (x2) =⇒ f is increasing on (a, b).
(c) If f0(x) < 0 for all x ∈ (a, b), then f is decreasing on (a, b), i.e. f (x1) > f (x2) for any x1 < x2 ∈ (a, b).
Proof For any x1 < x2 ∈ (a, b), there exits c ∈ (x1, x2), by the Mean Value Theorem, such that
f (x1) − f (x2) = f0(c)(x1− x2) > 0 =⇒f (x1) > f (x2) =⇒ f is decreasing on (a, b).
(d) Let f be twice differentiable on I = (a, b). If f00(x) > 0 for all x ∈ I = (a, b), then f is concave upward on I, i.e. For each c ∈ I, the graph of y = f (x), for x near c, lies above the tangent line to y = f (x) at (c, f (c)).
Proof For each c ∈ I, since
f00(c) = lim
x→c
f0(x) − f0(c) x − c > 0, there exists δ > 0 such that if 0 < |x − c| < δ then
f0(x) − f0(c) x − c > 0
=⇒
(f0(x) − f0(c) > 0 ∀ x ∈ (c, c + δ) f0(x) − f0(c) < 0 ∀ x ∈ (c − δ, c).
This imples that if 0 < |x − c| < δ, by the Mean Value Theorem, then f (x) − f (c) − f0(c)(x − c)
= [f0(z) − f0(c)] (x − c) for some z lying between x and c
> 0
This proves that the point (x, f (x)) in the graph of y = f (x) lies above the point (x, f (c) + f0(c)(x − c)) in the tangent line to the graph of y = f (x) at (c, f (c)).
(e) Let f be twice differentiable on I = (a, b). If f00(x) < 0 for all x ∈ I = (a, b), then f is concave downward on I, i.e. For each c ∈ I, the graph of y = f (x), for x near c, lies below the tangent line to y = f (x) at (c, f (c)).
Cauchy Mean Value Theorem Let f, g be continuous on I = [a, b] and let f, g be differen- tiable on (a, b). Then there is a number c ∈ (a, b) such that
f0(c) [g(b) − g(a)] = g0(c)[f (b) − f (a)].
Proof
Case 1 If g(b) = g(a) then there exists c ∈ (a, b), by the Rolle’s Theorem, such that g0(c) = 0.
Hence, the equality holds.
Case 2 If g(b) 6= g(a) the function h defined on I by h(x) = f (x) − f (a) −f (b) − f (a)
g(b) − g(a)[g(x) − g(a)] if x ∈ [a, b].
Since h is continuous on I = [a, b], differentiable on (a, b) and satisfies that h(a) = h(b) = 0, there exists c ∈ (a, b), by the Rolle’s Theorem, such that
0 = h0(c) = f0(c) − f (b) − f (a)
g(b) − g(a) g0(c) =⇒ f0(c) [g(b) − g(a)] = g0[f (b) − f (a)].
An Indeterminate Form 00 and l’Hˆopital’s Rule Let f, g be continuous on I = [a, b] and let f, g be differentiable on (a, b). Suppose that
(a) f (c) = 0 = g(c) for some c ∈ (a, b), i.e. lim
x→cf (x) = 0 = lim
x→cg(x), (b) g0(x) 6= 0 for all x ∈ (a, c) ∪ (c, b),
(c) lim
x→c
f0(x)
g0(x) = L ∈ R exists.
Then
limx→c
f (x)
g(x) = lim
x→c
f0(x) g0(x) = L.
Outline of the Proof
x→clim f (x)
g(x) = lim
x→c
f (x) − f (c)
g(x) − g(c) since f (c) = g(c) = 0
= lim
x→c
f0(t)
g0(t) for some t lying between x and c by the Cauchy Mean Value Theorem
= lim
t→c
f0(t)
g0(t) by observng that t → c whenever x → c
= L.
An Indeterminate Form ±∞∞ and l’Hˆopital’s Rule Let c be a point in the interval (a, b) and let f, g be be differentiable on (a, c) ∪ (c, b). Suppose that
(a) lim
x→cf (x) = ±∞ and lim
x→cg(x) = ±∞, (b) g0(x) 6= 0 for all x ∈ (a, c) ∪ (c, b),
(c) lim
x→c
f0(x)
g0(x) = L ∈ R exists.
Then
limx→c
f (x)
g(x) = lim
x→c
f0(x) g0(x) = L.
Outline of the Proof Case 1 Suppose that lim
x→c
f0(x)
g0(x) = L = 0.
Given > 0, since lim
x→c
f0(x)
g0(x) = L = 0, there exists δ1 > 0 such that if 0 < |x − c| < δ1 then
f0(x) g0(x)
< .
For any points x and x1 satisfying
either c < x < x1 < c + δ1, or c − δ1 < x1 < x < c
there exists x2 ∈ (c − δ1, c) ∪ (c, c + δ1) lying between x and x1, by the Cauchy Mean Value Theorem, such that
f (x) − f (x1) g(x) − g(x1)
=
f0(x2) g0(x2)
< .
Also, since
limx→c
1 −f (x1) f (x) 1 −g(x1)
g(x)
= 1,
there exists δ2 > 0 such that
if 0 < |x − c| < δ2 then
1 − f (x1) f (x) 1 −g(x1)
g(x)
> 1 2.
Let δ = min{δ1, δ2} such that if 0 < |x − c| < δ then
1 2
f (x) g(x)
<
f (x) g(x)
1 −f (x1) f (x) 1 −g(x1)
g(x)
=
f (x) − f (x1) g(x) − g(x1)
=
f0(x2) g0(x2)
< which implies that
f (x) g(x)
< 2.
Hence,
limx→c
f (x)
g(x) = 0 = L = lim
x→c
f0(x) g0(x). Case 2 Suppose that lim
x→c
f0(x)
g0(x) = L 6= 0.
Consider the function h defined by h(x) = f (x) − L g(x) for all x ∈ (a, c) ∪ (c, b) and note that limx→c
h0(x) g0(x) = lim
x→c
f0(x)
g0(x) − L = 0.
Applying Case 1, we get limx→c
f (x)
g(x) − L = lim
x→c
h(x)
g(x) = 0 =⇒ lim
x→c
f (x) g(x) = L.
