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 f^{0}(c) = 0 or f^{0}(c) does not exist.

Extreme Value Theorem Let f be continuous on [a, b]. Then there exist x_{1}, x_{2} ∈ [a, b] such
that

min

[a,b] f = f (x_{1}) ≤ f (x) ≤ f (x_{2}) = 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 f^{0}(c) exists, then f^{0}(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 f^{0}(c) = 0.

Proof Suppose that

lim

h→0

f (c + h) − f (c)

h = f^{0}(c) > 0,

since = f^{0}(c)

2 > 0, there exists δ > 0 such that if 0 < |h| < δ then

f (c + h) − f (c)

h − f^{0}(c)

< f^{0}(c)

2 ⇔ f^{0}(c)

2 > f (c + h) − f (c)

h − f^{0}(c) > −f^{0}(c)
2

=⇒ f (c + h) − f (c)

h > f^{0}(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 f^{0}(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) thenf^{0}(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 f^{0}(c) = 0.

Proof

Case 1 Suppose that f (x) = f (a) = f (b) for all x ∈ [a, b] =⇒ f^{0}(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, f^{0}(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

f^{0}(c) = f (b) − f (a)

b − a ⇐⇒ f (b) − f (a) = f^{0}(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 = g^{0}(c) = f^{0}(c) − f (b) − f (a)

b − a =⇒ f^{0}(c) = f (b) − f (a)
b − a .

Remarks Let f be differentiable on (a, b). Suppose that f^{0}(x) 6= 0 for all x ∈ (a, b). Then
(a) f is 1 − 1 on (a, b).

Proof For any x_{1} 6= x_{2} ∈ (a, b), there exits c lying between x_{1} and x_{2}, by the Mean Value
Theorem, such that

f (x1) − f (x2) = f^{0}(c)(x1− x2) 6= 0 =⇒f (x1) 6= f (x2) =⇒ f is 1 − 1 on (a, b).

(b) If f^{0}(x) > 0 for all x ∈ (a, b), then f is increasing on (a, b), i.e. f (x_{1}) < f (x_{2}) for any
x_{1} < x_{2} ∈ (a, b).

Proof For any x_{1} < x_{2} ∈ (a, b), there exits c ∈ (x_{1}, x_{2}), by the Mean Value Theorem, such
that

f (x_{1}) − f (x_{2}) = f^{0}(c)(x_{1}− x_{2}) < 0 =⇒f (x_{1}) < f (x_{2}) =⇒ f is increasing on (a, b).

(c) If f^{0}(x) < 0 for all x ∈ (a, b), then f is decreasing on (a, b), i.e. f (x_{1}) > f (x_{2}) 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 (x_{1}) − f (x_{2}) = f^{0}(c)(x_{1}− x_{2}) > 0 =⇒f (x_{1}) > f (x_{2}) =⇒ f is decreasing on (a, b).

(d) Let f be twice differentiable on I = (a, b). If f^{00}(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

f^{00}(c) = lim

x→c

f^{0}(x) − f^{0}(c)
x − c > 0,
there exists δ > 0 such that if 0 < |x − c| < δ then

f^{0}(x) − f^{0}(c)
x − c > 0

=⇒

(f^{0}(x) − f^{0}(c) > 0 ∀ x ∈ (c, c + δ)
f^{0}(x) − f^{0}(c) < 0 ∀ x ∈ (c − δ, c).

This imples that if 0 < |x − c| < δ, by the Mean Value Theorem, then
f (x) − f (c) − f^{0}(c)(x − c)

= [f^{0}(z) − f^{0}(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) +
f^{0}(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 f^{00}(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

f^{0}(c) [g(b) − g(a)] = g^{0}(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 g^{0}(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 = h^{0}(c) = f^{0}(c) − f (b) − f (a)

g(b) − g(a) g^{0}(c) =⇒ f^{0}(c) [g(b) − g(a)] = g^{0}[f (b) − f (a)].

An Indeterminate Form ^{0}_{0} 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) g^{0}(x) 6= 0 for all x ∈ (a, c) ∪ (c, b),

(c) lim

x→c

f^{0}(x)

g^{0}(x) = L ∈ R exists.

Then

limx→c

f (x)

g(x) = lim

x→c

f^{0}(x)
g^{0}(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

f^{0}(t)

g^{0}(t) for some t lying between x and c by the Cauchy Mean Value Theorem

= lim

t→c

f^{0}(t)

g^{0}(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) g^{0}(x) 6= 0 for all x ∈ (a, c) ∪ (c, b),

(c) lim

x→c

f^{0}(x)

g^{0}(x) = L ∈ R exists.

Then

limx→c

f (x)

g(x) = lim

x→c

f^{0}(x)
g^{0}(x) = L.

Outline of the Proof Case 1 Suppose that lim

x→c

f^{0}(x)

g^{0}(x) = L = 0.

Given > 0, since lim

x→c

f^{0}(x)

g^{0}(x) = L = 0, there exists δ_{1} > 0 such that
if 0 < |x − c| < δ_{1} then

f^{0}(x)
g^{0}(x)

< .

For any points x and x_{1} satisfying

either c < x < x_{1} < c + δ_{1}, or c − δ_{1} < x_{1} < x < c

there exists x_{2} ∈ (c − δ_{1}, c) ∪ (c, c + δ_{1}) lying between x and x_{1}, by the Cauchy Mean Value
Theorem, such that

f (x) − f (x1)
g(x) − g(x_{1})

=

f^{0}(x2)
g^{0}(x_{2})

< .

Also, since

limx→c

1 −f (x_{1})
f (x)
1 −g(x_{1})

g(x)

= 1,

there exists δ_{2} > 0 such that

if 0 < |x − c| < δ_{2} then

1 − f (x_{1})
f (x)
1 −g(x_{1})

g(x)

> 1 2.

Let δ = min{δ_{1}, δ_{2}} such that if 0 < |x − c| < δ then

1 2

f (x) g(x)

<

f (x) g(x)

1 −f (x_{1})
f (x)
1 −g(x_{1})

g(x)

=

f (x) − f (x_{1})
g(x) − g(x_{1})

=

f^{0}(x_{2})
g^{0}(x_{2})

< which implies that

f (x) g(x)

< 2.

Hence,

limx→c

f (x)

g(x) = 0 = L = lim

x→c

f^{0}(x)
g^{0}(x).
Case 2 Suppose that lim

x→c

f^{0}(x)

g^{0}(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

h^{0}(x)
g^{0}(x) = lim

x→c

f^{0}(x)

g^{0}(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.