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November 25, 2019 MAT186 – Week 8 Justin Ko
1 Mean Value Theorem
Theorem1(MeanValueTheorem). Let f be a continuous on [a,b] and differentiable on (a,b). There
exists a c ∈ (a,b) such that f(b)−f(a)
f′(c) = b −a .
Example 1: The Mean Value Theorem says there exists at least one point c ∈ (a,b) such that the
tangent line to the curve at (c,f(c)) is parallel to the secant line connecting (a,f(a)) and (b,f(b)).
40
30
20
10
−4 −3 −2 −1 1 2 3 4 5 6
−10
−20
Remark. Ifwealsoassumethatf(a) = f(b), thenthemeanvaluetheoremsaysthereexistsac ∈ [a,b]
such that f′(c) = 0. This result is called Rolle’s Theorem.
1.1 Consequences of the Mean Value Theorem
Corollary 1. If f′(x) = 0 for all x ∈ (a,b), then f is constant on the interval (a,b).
Corollary 2. If f′(x) = g′(x) for all x ∈ (a,b), then there is a constant C such that for all x ∈ (a,b),
f(x) = g(x)+C.
Corollary 3. If f′(x) > 0 for all x ∈ (a,b), then f is strictly increasing on (a,b).
Corollary 4. If f′(x) < 0 for all x ∈ (a,b), then f is strictly decreasing on (a,b).
1.2 Example Problems
Problem 1.1. (⋆⋆) Suppose that f(x) is an even function, i.e. f(−x) = f(x). Prove that its integral
F(x) = Rxf(t)dt is odd.
0
Solution 1.1. It suffices to show that F(−x) = −F(x), that is g(x) := F(x) + F(−x) = 0. By the
chain rule and the fundamental theorem of calculus
g′(x) = d (F(x)+F(−x)) = F′(x)+F′(−x) = f(x)−f(−x) = f(x)−f(x) = 0,
dx
since f(x) is even. Since g′(x) = 0, the Corollary 1 implies that g(x) is constant. To find out what
this constant is, we see that g(0) = F(0) + F(−0) = 0 so the constant must be 0. In particular,
F(x)+F(−x)=0for all x, which implies that F(x) is an odd function.
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November 25, 2019 MAT186 – Week 8 Justin Ko
x
Problem 1.2. (⋆⋆) Show that 1+x < e for all x > 0.
x x ′ x
Solution 1.2. Considerthefunctionf(x) = 1+x−e . Wehavee > 1forx > 0, sof (x) = 1−e < 0
for all x > 0. Corollary 4 implies f is strictly decreasing for x > 0. Therefore, for x > 0, we have
x x
0 = f(0) > f(x) = 1+x−e =⇒ 1+x0.
Alternate Method: Let x > 0. Notice that the function et is continuous on [0,x] and differentiable
on (0,x), so the mean value theorem states there exists a c ∈ (0,x) such that
x 0 x
f′(c) = e −e = e −1.
x−0 x
′ c
Furthermore, we have f (c) = e > 1 since c > 0. Therefore, the equation above implies
ex −1
′ x x
x =f (c) > 1 =⇒ e −1>x =⇒ e >1+x.
Problem 1.3. (⋆⋆⋆) Use the Mean Value Theorem to prove Corollary 1.
Solution 1.3. Suppose that f′(x) = 0 for all x ∈ (a,b). Consider the points a < x1 < x2 < b. Since
f′ exists on (a,b), we have f is continuous on [x ,x ] and differentiable on (x ,x ). Therefore, by the
1 2 1 2
Mean Value Theorem, there exists a c ∈ (x ,x ) such that
1 2
f(x )−f(x )
f′(c) = 2 1 .
x −x
2 1
By assumption, we must have f′(c) = 0, which implies that
f(x2)−f(x1) = 0 =⇒ f(x ) = f(x ).
x −x 1 2
2 1
Since x and x were arbitrary, we can conclude that f(x ) = f(x ) for all x ,x ∈ (a,b). That is,
1 2 1 2 1 2
f(x) is constant.
Problem 1.4. (⋆⋆⋆) Prove Corollary 2.
Solution 1.4. Suppose that f′(x) = g′(x) for all x ∈ (a,b). Consider the function h(x) = f(x)−g(x).
′ ′ ′
By assumption we have h (x) = f (x) − g (x) = 0 for all x ∈ (a,b). Therefore, using Corollary 1, we
can conclude that there exists a constant C such that h(x) = C on (a,b). Therefore, we have
C=h(x)=f(x)−g(x) =⇒ f(x)=g(x)+C.
Problem 1.5. (⋆⋆⋆) Prove Corollary 3.
Solution 1.5. Suppose that f′(x) > 0 for all x ∈ (a,b). To show a function is strictly increasing, we
need to show that
x 0, which implies that
f(x2)−f(x1) > 0 =⇒ f(x ) > f(x ).
x −x 2 1
2 1
Remark. To prove Corollary 4, we can use the fact that f is strictly decreasing on (a,b) is equivalent
to −f is strictly increasing on (a,b). Corollary 4 then follows as a direct consequence of Corollary 3.
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November 25, 2019 MAT186 – Week 8 Justin Ko
2 Extreme Values
Definition 1. The largest and smallest values a function can take on an interval I are called the global
extreme values. There are two types:
1. f has a global maximum on I at c ∈ I if f(x) ≤ f(c) = M for all x ∈ I. The number M is called
the global maximum value of f. The point c is sometimes called the global maximizer.
2. f has a global minimum on I at c ∈ I if f(x) ≥ f(c) = m for all x ∈ I. The number m is called
the global minimum value of f. The point c is sometimes called the global minimizer.
Definition 2. The largest and smallest values a function can take near a point are called local extreme
values. There are two types:
1. f has a local maximum at c if there exists an open interval I containing c such that f(x) ≤ f(c)
for all x ∈ I.
2. f has a local minimum at c if there exists an open interval I containing c such that f(x) ≥ f(c)
for all x ∈ I.
3
Example 2: The function f(x) = x −12x on the interval [−3,4.5] has a local maximum at x = −2
and a local minimum at x = 2. The function has a global maximum value of f(4.5) = 37.125 at
x=4.5 and a global minimum value of f(2) = −16 at x = 2.
Example 3: The function f(x) = x on the interval (−3,3) does not have any extreme values. Even
though it looks like the the global minimums and maximums of f occur at −3 and 3, f(3) and f(−3)
do not exist since −3 and 3 are not in the interval (−3,3), so they cannot be extreme values.
40 5
f(x) = x
3
f(x) = x −12x 4
30
3
2
20
1
10
−4 −3 −2 −1 1 2 3 4
−1
−4 −3 −2 −1 1 2 3 4 5 6 −2
−10 −3
−4
−20 −5
2.1 Existence of Extreme Values
Some functions may not have extreme values. The following theorem provides some conditions that
guarantee the existence of extreme values.
Theorem 2 (Extreme Value Theorem). If f is continuous on a closed interval [a,b] then f has a
global maximum and a global minimum on [a,b]. That is, there exists points c and d in [a,b] such that
f(c) ≤ f(x) ≤ f(d) for all x ∈ [a,b].
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November 25, 2019 MAT186 – Week 8 Justin Ko
2.2 Location of Extreme values
Definition 3. The critical points of f are the values c in the domain of f such that f′(c) = 0 or f′(c)
does not exist.
Theorem 3. If f(x) has a local extreme at c, then c is a critical point of f.
This means we look for local extremes at the critical points of our function. However, this does not
mean that every critical point corresponds to a extreme value. Theorem 3 is insufficient to determine
if the critical point is a local minimum, local maximum, or neither.
2.3 Classification of Extreme values
Wecan use something called the first derivative test to classify continuous function at critical points.
Theorem 4 (First Derivative Test). Suppose that c is a critical point of a continuous function f.
1. If f′ changes signs from negative to positive at c, then f has a local minimum at c.
2. If f′ changes signs from positive to negative at c, then f has a local minimum at c.
If our function is twice differentiable, we can instead use something called the second derivative
test to classify critical points of nice twice differentiable functions.
Theorem 5 (Second Derivative Test). Suppose that f′′(x) is continuous near c.
1. If f′(c) = 0 and f′′(c) > 0, then f has a local minimum at c.
2. If f′(c) = 0 and f′′(c) < 0, then f has a local maximum at c.
To find the global minimum and maximum values of a function on an interval [a,b], it suffices to
evaluate our function at the critical points and the boundary of our interval and choose the corre-
sponding smallest and largest values.
2.4 Example Problems
x −x
Problem 2.1. (⋆⋆) Find all local and global extreme values of f(x) = e +e .
2
Solution 2.1.
Finding the Critical Points: We first find the critical points of our function. Using the chain
rule, we see
x −x
f′(x) = e −e .
2
The derivative exists for all x ∈ R, so we can set this equal to 0 to find the remaining critical points,
x −x
f′(x) = e −e =0⇒ex=e−x⇒x=−x⇒x=0.
2
Classifying the Critical Points: To classify the critical point at x = 0, notice that
x −x
f′′(0) = e +e =1>0,
2
x=0
so there is a local minimum when x = 0. The minimum value at this local minimum is f(0) = 0.
Wenowsearch for global maximum and minimum values. Taking x → ±∞, we see that
x −x x −x
lim e +e =∞ and lim e +e =∞
x→∞ 2 x→−∞ 2
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