Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
2008
Ch4-1: Local Extrema and The Mean Value Theorem
Definition
1 A function f defined on a set D has a
local(relative) maximum at a point c if∃δ >0 s.t.
f(c)≥f(x) ∀x ∈(c−δ,c+δ)∩D
2 A function f defined on a set D has a
local(relative) minimum at a point c if∃δ >0 s.t.
f(c)≤f(x) ∀x ∈(c−δ,c+δ)∩D Ex1: Find all local extrema and the global extrema.
1 f(x) =|x2−1|,−4≤x ≤6
2 f(x) =x(x −1), x ∈[0,1]
WEN-CHINGLIEN Calculus (I)
Ch4-1: Local Extrema and The Mean Value Theorem
Definition
1 A function f defined on a set D has a
local(relative) maximum at a point c if∃δ >0 s.t.
f(c)≥f(x) ∀x ∈(c−δ,c+δ)∩D
2 A function f defined on a set D has a
local(relative) minimum at a point c if∃δ >0 s.t.
f(c)≤f(x) ∀x ∈(c−δ,c+δ)∩D Ex1: Find all local extrema and the global extrema.
1 f(x) =|x2−1|,−4≤x ≤6
2 f(x) =x(x −1), x ∈[0,1]
Ch4-1: Local Extrema and The Mean Value Theorem
Definition
1 A function f defined on a set D has a
local(relative) maximum at a point c if∃δ >0 s.t.
f(c)≥f(x) ∀x ∈(c−δ,c+δ)∩D
2 A function f defined on a set D has a
local(relative) minimum at a point c if∃δ >0 s.t.
f(c)≤f(x) ∀x ∈(c−δ,c+δ)∩D Ex1: Find all local extrema and the global extrema.
1 f(x) =|x2−1|,−4≤x ≤6
2 f(x) =x(x −1), x ∈[0,1]
WEN-CHINGLIEN Calculus (I)
Ch4-1: Local Extrema and The Mean Value Theorem
Definition
1 A function f defined on a set D has a
local(relative) maximum at a point c if∃δ >0 s.t.
f(c)≥f(x) ∀x ∈(c−δ,c+δ)∩D
2 A function f defined on a set D has a
local(relative) minimum at a point c if∃δ >0 s.t.
f(c)≤f(x) ∀x ∈(c−δ,c+δ)∩D Ex1: Find all local extrema and the global extrema.
1 f(x) =|x2−1|,−4≤x ≤6
2 f(x) =x(x −1), x ∈[0,1]
Ch4-1: Local Extrema and The Mean Value Theorem
Definition
1 A function f defined on a set D has a
local(relative) maximum at a point c if∃δ >0 s.t.
f(c)≥f(x) ∀x ∈(c−δ,c+δ)∩D
2 A function f defined on a set D has a
local(relative) minimum at a point c if∃δ >0 s.t.
f(c)≤f(x) ∀x ∈(c−δ,c+δ)∩D Ex1: Find all local extrema and the global extrema.
1 f(x) =|x2−1|,−4≤x ≤6
2 f(x) =x(x −1), x ∈[0,1]
WEN-CHINGLIEN Calculus (I)
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
WEN-CHINGLIEN Calculus (I)
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
WEN-CHINGLIEN Calculus (I)
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
WEN-CHINGLIEN Calculus (I)
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
Theorem (Fermat’s Theorem)
If f has a local extrema at an interior point c and f′(c) exist, then f′(c) = 0.
Proof.
f′(c) = lim
x→c
f(x)−f(c) x −c
Suppose that f has a local maximum at an interior point c,
⇒ ∃δ >0 s.t.
f(x)≤f(c),∀x ∈(c−δ,c+δ)
⇒ lim
x→c−
f(x)−f(c)
x−c ≥0
On the other hand, lim
x→c+
f(x)−f(c)
x−c ≤0
⇒ f′(c) =0
WEN-CHINGLIEN Calculus (I)
Definition (Absolute Extreme Value) Let f(x)be defined on the domain D
1 f has an absolute maximum at d if f(d)≥f(x),∀x ∈D
2 f has an absolute minimum at d if f(d)≤f(x),∀x ∈D Ex1:
1 f(x) =x3
2 f(x) =|x|(Differentiability)
Definition (Absolute Extreme Value) Let f(x)be defined on the domain D
1 f has an absolute maximum at d if f(d)≥f(x),∀x ∈D
2 f has an absolute minimum at d if f(d)≤f(x),∀x ∈D Ex1:
1 f(x) =x3
2 f(x) =|x|(Differentiability)
WEN-CHINGLIEN Calculus (I)
Definition (Absolute Extreme Value) Let f(x)be defined on the domain D
1 f has an absolute maximum at d if f(d)≥f(x),∀x ∈D
2 f has an absolute minimum at d if f(d)≤f(x),∀x ∈D Ex1:
1 f(x) =x3
2 f(x) =|x|(Differentiability)
Definition (Absolute Extreme Value) Let f(x)be defined on the domain D
1 f has an absolute maximum at d if f(d)≥f(x),∀x ∈D
2 f has an absolute minimum at d if f(d)≤f(x),∀x ∈D Ex1:
1 f(x) =x3
2 f(x) =|x|(Differentiability)
WEN-CHINGLIEN Calculus (I)
Definition (Absolute Extreme Value) Let f(x)be defined on the domain D
1 f has an absolute maximum at d if f(d)≥f(x),∀x ∈D
2 f has an absolute minimum at d if f(d)≤f(x),∀x ∈D Ex1:
1 f(x) =x3
2 f(x) =|x|(Differentiability)
Theorem (M.V.T)
If f is continuous on[a,b]and differentiable on(a,b), then∃c ∈(a,b)s.t.
f(b)−f(a)
b−a =f′(c) Theorem (Rolle’s Theorem)
If f is continuous on[a,b]and differentiable on(a,b)with f(a) =f(b), then there exists a number c ∈(a,b)s.t.
f′(c) =0.
WEN-CHINGLIEN Calculus (I)
Theorem (M.V.T)
If f is continuous on[a,b]and differentiable on(a,b), then∃c ∈(a,b)s.t.
f(b)−f(a)
b−a =f′(c) Theorem (Rolle’s Theorem)
If f is continuous on[a,b]and differentiable on(a,b)with f(a) =f(b), then there exists a number c ∈(a,b)s.t.
f′(c) =0.
Theorem (M.V.T)
If f is continuous on[a,b]and differentiable on(a,b), then∃c ∈(a,b)s.t.
f(b)−f(a)
b−a =f′(c) Theorem (Rolle’s Theorem)
If f is continuous on[a,b]and differentiable on(a,b)with f(a) =f(b), then there exists a number c ∈(a,b)s.t.
f′(c) =0.
WEN-CHINGLIEN Calculus (I)
Corollary
If f is continuous on[a,b]and differentiable on(a,b), s.t.
m ≤f′(x)≤M,∀x ∈(a,b),
then m(b−a)≤f(b)−f(a)≤M(b−a).
Corollary
If f is continuous on[a,b]and differentiable on(a,b)with f′(x) =0∀x ∈(a,b), then f is constant on(a,b).
Ex2: Assume that f(x) = −x2+2. Explain why∃a point c ∈(−1,2)s.t. f′(c) =−1
Ex3: Show that |cos x−cos y| ≤ |x −y|.
Corollary
If f is continuous on[a,b]and differentiable on(a,b), s.t.
m ≤f′(x)≤M,∀x ∈(a,b),
then m(b−a)≤f(b)−f(a)≤M(b−a).
Corollary
If f is continuous on[a,b]and differentiable on(a,b)with f′(x) =0∀x ∈(a,b), then f is constant on(a,b).
Ex2: Assume that f(x) = −x2+2. Explain why∃a point c ∈(−1,2)s.t. f′(c) =−1
Ex3: Show that |cos x−cos y| ≤ |x −y|.
WEN-CHINGLIEN Calculus (I)
Corollary
If f is continuous on[a,b]and differentiable on(a,b), s.t.
m ≤f′(x)≤M,∀x ∈(a,b),
then m(b−a)≤f(b)−f(a)≤M(b−a).
Corollary
If f is continuous on[a,b]and differentiable on(a,b)with f′(x) =0∀x ∈(a,b), then f is constant on(a,b).
Ex2: Assume that f(x) = −x2+2. Explain why∃a point c ∈(−1,2)s.t. f′(c) =−1
Ex3: Show that |cos x−cos y| ≤ |x −y|.
Corollary
If f is continuous on[a,b]and differentiable on(a,b), s.t.
m ≤f′(x)≤M,∀x ∈(a,b),
then m(b−a)≤f(b)−f(a)≤M(b−a).
Corollary
If f is continuous on[a,b]and differentiable on(a,b)with f′(x) =0∀x ∈(a,b), then f is constant on(a,b).
Ex2: Assume that f(x) = −x2+2. Explain why∃a point c ∈(−1,2)s.t. f′(c) =−1
Ex3: Show that |cos x−cos y| ≤ |x −y|.
WEN-CHINGLIEN Calculus (I)
Corollary
If f is continuous on[a,b]and differentiable on(a,b), s.t.
m ≤f′(x)≤M,∀x ∈(a,b),
then m(b−a)≤f(b)−f(a)≤M(b−a).
Corollary
If f is continuous on[a,b]and differentiable on(a,b)with f′(x) =0∀x ∈(a,b), then f is constant on(a,b).
Ex2: Assume that f(x) = −x2+2. Explain why∃a point c ∈(−1,2)s.t. f′(c) =−1
Ex3: Show that |cos x−cos y| ≤ |x −y|.
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
WEN-CHINGLIEN Calculus (I)
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
WEN-CHINGLIEN Calculus (I)
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
WEN-CHINGLIEN Calculus (I)
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
WEN-CHINGLIEN Calculus (I)
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
Proof of Rolle’s Theorem.
Case1: If f(x) = constant=⇒f′(x) =0.
Case2: (Suppose that) f(x)is not a constant function.
Since f(x)is continuous on[a,b], so max
x∈[a,b]f(x)and min
x∈[a,b]f(x)are achieved.
Therefore, there exists x0 ∈(a,b) such that f(x0)>f(a)or f(x0)<f(a).
⇒ A global extremum, say, f(c), occurs with c∈(a,b).
⇒ By Fermat’s Theorem, f′(c) = 0.
WEN-CHINGLIEN Calculus (I)
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
WEN-CHINGLIEN Calculus (I)
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
WEN-CHINGLIEN Calculus (I)
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
Proof of M.V.T.
Define F(x) =f(x)− f(b)b−a−f(a)(x−a), continuous and differentiable
⇒F(a) =f(a) =F(b)
By Rolle’s theorem, ∃c ∈(a,b)s.t. F′(c) =0
⇒F′(c) =f′(c)− f(b)−f(a) b−a
⇒f′(c) = f(b)−f(a) b−a
WEN-CHINGLIEN Calculus (I)
