2011s-calc1-lec-4-1

8/12/2019 2011S-Calc1-Lec-4-1

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4.1. Maximum and Minimum Values

1. Denition (maximum, minimum, local maximum, and local minimum)

Assume that x = c is in the domain of a function f (x).

• We say f (c) is a maximum(also called as an absolute maximum or a global maximum) if f (c) ≥ f (x) for

all x values in the domain.

• We say f (c) is minimum(also called as an absolute minimum or a global minimum) if f (c) ≤f (x) for allx values in the domain.

• We say f (c) is a local maximum(relative maximum) if c is not an endpoint of the domain and f (c) is amaximum in a small neighborhood of c.

• We say f (c) is a local minimum(relative minimum) if c is not an endpoint of the domain and f (c) is aminimum in a small neighborhood of c.

2. Example Find any maxima, minima, local maxima, and local minima of the function whose graph is shown :

(a) f (x) is dened on the domain [0, 6] (b) f (x) is dened on the domain [−1, 7]

Answers :

(a) Maximum does not exist, minimum − 2 , local minimum 2, and local maximum does not exist.(b) Max 6, minimum does not exist, local maximum 6 and 2, and local minimum 0.

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3. Theorem (Extreme value theorem) A continuous function on a closed interval [a, b] has always global maximumand global minimum.

Notice that a discontinuous function or a function whose domain is an open interval may fail to have extrema.

4. Denition (critical number) x = c is called a critical number of the function y = f (x) if c is in the domain of thefunction and either

• f (xo) = 0

or

• the derivative does not exist when x = xo .

(Various cases of critical numbers)

Note that if f (xo) is a local max/min, then xo is a critical value. The following theorem states this.However, be warned that not all critical numbers are local maximum or minimum. For example, (0, 0) isa critical point of the function f (x) = x3 because f (0) = 0 , but obviously (0, 0) is not a local maximumor minimum.

5. Theorem (Fermat’s theorem for critical numbers)If a function f (x) has a local maximum or local minimum at x = xo and f (xo) exists, then f (xo) = 0 .

Proof : Suppose that f (x o ) is a local maximum. Then f (x o ) ≥ f (x ) for all x in a small neighborhood of x o . Therefore

limh → 0+

f (x o + h ) − f (x o )h

≤ 0 because f (x o + h ) ≤ f (x o ) and h > 0

limh → 0 −

f (x o + h ) − f (x o )h

≥ 0 because f (x o + h ) ≤ f (x o ) and h < 0

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Because f (x o ) exists, above two limits should give the same value. Hence f (x o ) = 0 .

If f (x o ) is a local minimum, let g(x ) = − f (x ) so that g(x o ) is a local maximum of g(x ) . By the previous proof, g (x o ) = − f (x o ) = 0 , and

therefore, f (x o ) = 0 , again.

6. Example Find critical numbers of the function :

(a) f (x) = x3 + 3 x2 −24x

(b) g(x) = |x −1|+ 3(c) h(x) = 3√ 4 −x2

Answers : (a) x = − 4, 2 (b) x = 1 (c) x = 0 , ± 2

7, Property (Closed interval method)A continuous function y = f (x) on a closed interval [a, b] always has the global maximum and minimumamong

”critical points” and ”end-points”

8. Example Find the maximum and minimum values of f on the given interval :

(a) f (x) = x3 −3x + 1 , I = [0, 3]

(b) g(x) = x2 −4x2 + 4

, I = [−4, 4]

(c) T (θ) = 2cos θ + sin 2 θ, I = [0, 2π]

Answers :

(a) Max f (3) = 19 , min f (1) = − 1(b) Max g(4) = f (− 4) = 0 .6, min g(0) = − 1(c) Max T (0) = T (2 π ) = 2 , min T (π ) = − 2

9. Example (# 64 in textbook p212) An object with weight W is dragged along a horizontal plane by a force actingalong a rope attached to the object. If the rope makes an angle θ with the plane, then the magnitude of the force is

F = µW µ sin θ + cos θ

where µ is a positive constant between 0 and 1 called the coefcient of friction and where 0 ≤ θ ≤ π/ 2.Show that F is minimized when tan θ = µ.

(Answer) We look for the minimum of F (θ) while 0 ≤ θ ≤ π/ 2. For this, we rst look for critical numbers of F .

F (θ) = − µW (µ cos θ − sin θ)

(µ sin θ + cos θ)2

and F = 0 when the numerator − µW (µ cos θ − sin θ) = 0 . If we denote the critical number of the function by c,

µ cos c − sin c = 0 ; µ cos c = sin c ; µ = sin ccos c

= tan c

That is, the critical number is c = tan − 1 (µ) . Notice

F (0) = µW = sin ccos c

W

F (π/ 2) = W

F (c) = µW

µ sin c + cos c=

sin ccos c · W

sin ccos c · sin c + cos c

= sin c · W

sin 2 c + cos 2 c= sin c · W

.

Since 0 ≤ µ ≤ 1, F (0) is smaller than F (π/ 2) . Also since 0 ≤ cos c ≤ 1, sin c · W ≤sin ccos c

W . That is, F (c) ≤ F (0) ≤ F (π/ 2) . F is thesmallest when θ = c and for this θ , µ = tan θ .

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2011s-calc1-lec-4-1

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