MATH 151 Engineering Math I, Spring 2014
JD Kim
Week12 Section 5.2, 5.3, 5.5
Section 5.2 Maximum and Minimum Values
Definition Maximum, Minimum
1. A function f has an Absolute Maximum at c if f (c) ≥ f (x) for all x in the
domain of f . A function f has an Absolute Minimum at c if f (c) ≤ f (x)
for all x in the domain of f . In this case, we call f (c) the maximum value or
minimum value, respectively.
2. A function f has an Local Maximum at c if f (c) ≥ f (x) when x is near c.
A function f has an Local Minimum at c if f (c) ≤ f (x) when x is near c.
Absolute maximum/minimum give y value only,
Local maximum/minimum give the point.
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Ex1) Find all absolute and local extrema by graphing the function:
1-1) f (x) = 1 − x2 , − 3 < x ≤ 2
1-2) f (x) =
1-3) f (x) =
1
, 0<x≤1
x



 x2


 2 − x2
if − 1 ≤ x < 0
if 0 ≤ x ≤ 1
2
Definition Critical Number
We call x = c a Critical Number of f (x) if x = c is in the domain of f and
either f ′ (c) = 0 or f ′ (c) does not exist.
Ex2) Find all critical values for following functions:
2-1) f (x) = 4x3 − 9x2 − 12x + 3
2-2) f (x) = |x2 − 1|
3
2-3) f (x) = x4/5 (x − 2)2
2-4) f (x) = xe2x
2-5) f (x) = x ln x
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Extreme Value Theorem
If f (x) is a continuous function on a close interval [a, b], then f will attain
both an absolute maximum and an absolute minimum.
Ex3) Use the extreme value theorem to find the absolute extrema for f (x) =
1 + 27x − x3 on the interval [0, 4].
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Ex4) Use the extreme value theorem to find the absolute extrema for f (x) =
x − 2 cos x on the interval [0, π].
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Section 5.3 Derivatives and Shapes of Curves
Mean Value Theorem (MVT)
If f (x) is continuous on the interval [a, b] and differentiable on the interval (a, b),
f (b) − f (a)
.
then there exists a number c, where a < c < b, so that f ′ (c) =
b−a
Graphically, this means the tangent line to the graph of f (x) at x = c is parallel to
the secent line joining the points (a, f (a)) and (b, f (b)).
Ex5) Given f (x) = 4 − x2 , show f (x) satisfies the mean value theorem on the
interval [1, 2] and find all c that satisfy the conclusion of the mean value theorem.
Ex6) Suppose 1 ≤ f ′ (x) ≤ 4 for all x in interval [2, 5]. Prove 3 ≤ f (5)−f (2) ≤ 12.
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First Derivativ Test for increasing/decreasing and local extrema
1. If f ′ > 0 on an interval I, then f is increasing on I.
2. If f ′ < 0 on an interval I, then f is decreasing on I.
3. If f ′ goes from positive to negative at x = a, and x = a is in the domain of f ,
then f has a local maximum at x = a.
4. If f ′ goes from negative to positive at x = a, and x = a is in the domain of f ,
then f has a local minimum at x = a.
Ex7) Find all intervals of increase and decrease and identify all local extrema.
7-1) f (x) = x4 + 4x3
√
7-2) f (x) = x x + 1
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7-3) f (x) = xe2x
Second Derivative Test for concavity and inflection points
1. If f ′′ > 0 on an interval I, then f ′ is increasing, hence f is concave upward on
I.
2. If f ′′ < 0 on an interval I, then f ′ is decreasing, hence f is concave downward
on I.
3. If f changes concavity at x = a, and x = a is in the domain of f , then x = a
is a inflection point of f .
Ex8) Find intervals of concavity and inflection points of we are given that f ′ (x) =
4x − 6x + 6.
3
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Ex9) Sketch the graph of f (x) = x ln x by locating intervals of increase/decrease,
local extrema, concavity, and inflection points.
10
x
by locating intervals of increase/decrease,
(x − 1)2
local extrema, concavity, and inflection points.
−x − 1
2x + 4
′′
For your convenience, f ′ (x) =
,
f
(x)
=
.
(x − 1)3
(x − 1)4
Ex10) Sketch the graph of f (x) =
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Second Derivative Test for local extreme
If x = c is a critical number for f (x), then
1. If f ′′ (c) > 0, then f is concave up, therefore f (x) has a local minimum at x = c.
2. If f ′′ (c) < 0, then f is concave down, therefore f (x) has a local maximum at
x = c.
3. If f ′′ (c) = 0 or does not exist, then the test fails, therefore use the first derivative
test to find the local extreme.
Ex11) Use the second derivative test to find the local extrema for f (x) = x3 − 3x − 1.
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Section 5.5 Applied Maximum and Minimum problem.
Steps in solving Applied Maximum and Minimum problems
1. Understand the problem
2. Draw a diagram
3. Introduce notation (Let Q is to be maximized or minimized)
4. Express Q in terms of some of other symbols from Step 3.
5. If Q has been expressed as a function of more than one variable in Step 4, use the
given information to find relationships (in the form of equations) among these
variables. Then use these equations to eliminate all but one of the variables in
the expression for Q. Thus Q will be given as a function of one variable x, say,
Q = f (x). Write the domain of this function.
6. Find the absolute maximum or minimum value of f .
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Ex12) A company wants to manufacture a box with a volume of 36 cubic feet.
The box has no top, and the length is twice of width. Find the dimensions of the
box that minimizes the amount of material used.
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Ex13) If 10,800 square centimeters of material is avaliable to make a box with a
square base and an open top, find the largest possible volume of the box.
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Ex14) A poster is to have an area of 180 square inches with 1-inch marginx at the
bottom and sides and 2-inch margin at top. what dimensions will give the largest
printed area?
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Ex15) Find the point on the line y = 2x − 3 that is closest to the point (−1, 3).
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Ex16) Find the dimensions of the rectangle of largest area that has its base on the
x axis and its other two vertices above the x axis lying on the parabola y = 8 − x2 .
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Ex17) Find the area of the largest rectangle that can be inscribed in a right
triangle with legs of length 3cm and 4cm if two sides of the rectangle lie along the
legs.
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Ex18) A piece of wire 10m long is cut into two pieces. One piece is bent into a
square and the other is bent into a circle. How should the wire be cut so that the
total enclosed area is a maximum? A minimum?
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