math 53 le 3 reviewer problems

7/25/2019 Math 53 LE 3 Reviewer problems

1/14

U P K E M

M E M B E R S H I P A C A D E M I C D E V E L O P M E N T

Math 53

3rd Long Exam Reviewer Problems

Worked out exercises from the Math 53 module

1 Reviewer

A. TRUE or FALSE

1. ( TRUE FALSE ) The graph of f(x) = g(x)

x2 4 where g(x) is a polynomial, has avertical asymptote atx = 2.

2. ( TRUE FALSE ) If the functionfhas a relative minimum at x = c, then f(c)> 0.

3. ( TRUE FALSE ) A point on the graph offcan be both a point of inflection and a

relative extremum off.

4. ( TRUE FALSE ) A function that has no absolute extremum value on [a, b] is discon-

tinuous on [a, b].

5. ( TRUE FALSE ) The function f(x) = 2x2 x +sin(2x) satisfies Rolles Theorem inthe interval

0, 1

2

.

B. 1. Givenf(x) = x3

x2 1 , f(x) =

x2(x2 3)(x2 1)2 , and f

(x) =2x(x2 + 3)

(x2 1)3 . (Note:

3 1.732.)

(a) Identify the domain, and the x- and y -intercept/s off.

(b) Using limits, find the equations of all the asymptotes of the graph off.

1


2/14

University of the Philippines Chemical Engineering Society, Inc. (UP KEM)

Math 53 - 3rd Long Exam Problem Set

(c) Determine the critical numbers and possible points of inflection off.

(d) Construct a table indicating the intervals wherefis increasing or decreasing and where its

is concave up or concave down, and the points where fhas relative extrema and points of

inflection.

(e) Sketch the graph off, with emphasis on linear asymptotes and concavity. Label all points

of interest.

Answer:

(a) dom(f) = (, 1) (1, 1) (1, +) or R/{1, 1}.x-intercept: 0

y-intercept: 0

(b) Vertical asymptotes are the easiest so lets find them first. By inspection, note that:

limx2

f(x) =

limx2

f(x) =

So two of our asymptotes are x = 2 and x= 2 .Next, try to verify that there are no horizontal asymptotes.

The degree of the numerator is greater than that of the denominator by exactly 1. So

we have an oblique asymptote. Proceed by expressing f(x) as q(x) + rd(x)

, where q ,r, d

are the quotient (polynomial), remainder and divisor (polynomial) respectively. In otherwords, do a long division.

f(x) = x3

x2 1

=x xx2 1 (see remark)

Now find a and b such that

0 = limx

x x

x2

1(ax+b)

= limx

[x (ax+b)]

0

limx

x

x2 1

= limx

x ax+b

Observe that as x gets arbitrarily large, the b term does not contribute to the limit, so

= limx

x ax

= limx

x(1 a)

Page 2 of 14


3/14



For the equation to hold true, (1a) must be equal to 0. Note that no matter how largex is, multiplying it by 0 results to 0

=

1

a= 0

= 1 =a

So we have y = x as an oblique asymptote. Sometimes, a shortcut for finding the

oblique asymptote isy = q(x) where qis the quotient obtained by the long division.

Remark. You are expected to already know how to do polynomial division.

(c) findc such thatf(c) = 0 or f(c) is undefined. Mentally solving for the zeroes, we have

x {0,

3, +

3}. For the denominator to be zero, we havex {1, 1}. The unionof these two sets is the required.

(d)

3 < 1 0 < 1 < 3f(x) 0 undef 0 undef 0

f(x) undef + 0 undef + +conclusion max v. asymp inf. pt. v. asymp min

(e) graph:

3, f(3

1

3, f(

3

1 x

f(x)

2. Do the above given: f(x) = (2x+ 1)(x+ 2)

(x+ 1)2 , f(x) =

1 x(x+ 1)3

, and f(x) = 2(x 2)(x+ 1)4

Page 3 of 14


4/14



Answer:

(a) dom(f) = (, 1) (1, +) or R/{1}.x-intercept: x=

2

x=

12

y-intercept: 2

(b) Vertical: x= 1Horizontal: y= 2

Oblique: none

(c) x {1, 1}

(d)

< 1 < 1 0 (13)

= minimum (14)

(7) (12) (14) = h= 1832

= 9 (15)

Therefore the dimensions required are r = 3 cm and r= 9 cm

4. Find two numbers that are both greater than or equal to 1 and whose sum is 10 such that the

product of one number and the fourth power of the other is maximum. How about if it is to be

minimum?

Answer: Let x, y 1 R be two, possibly distinct numbers. Find: max xy4, subject to

Page 8 of 14


9/14



the constraintx +y = 10. Then,

x= 10 y (1)

(1) =

max (10

y)y4= max 10y4 y5 (2)

Lety be the extremum.

= 0 = ddy

10y4 y5

y=y

(3)

(3) = 40y3

5y4

= 0 (4)

= (y = 0) (y = 8) (5)

With the second derivative test,

d2

dy2

10y4 y5y=0

= 0 (6)

= inconclusive (7)

d2

dy2

10y4 y5y=8

= 2560< 0 (8)

= maximum (9)(5) (9) = y = 8 (10)

(1) (10) = x = 2 (11)

Since 2 and 8 are both greater than or equal to 1, they satisfy the constraint. So the

maximum is 2 84 = 8192 occurring at (x, y) = (2, 8).

For the minimum, try and verify by yourself that it is 9 at (9, 1). Hint: Extreme Value

Theorem.

5. A rectangular box with an open top and a square base is to be made from 48 square feet of

material. What dimensions will result in a box with the largest possible volume?

Answer: Recall for a rectangular box:

Volume =lwh

Surface Area = 2(lw+lh+wh)

The box has an open top and a square base. Lets be the side length of the square base.

Page 9 of 14


10/14



Modifying the formula,

Volume =s2h

Surface Area = 4sh+s2

Our objective function is:

48 = 4sh+s2

= h= 48 s2

4s

= s2h= s(48 s2)

4

Following the usual steps, you should arrive at two candidate critical numbers, s {4, 4}.After applying the second derivative test, you should confirm that at s = 4, the second

derivative is less than zero. Thus a maximum. Actually, why would you even consider the

test? Lengths are always positive so you should discard s =4 immediately, right? No.Not all critical points are extremum, or even a maximum (for this particular case). Haha

Solve then for h. So the dimensions required are 4 4 8 ft3

6. Of the lines tangent to the graph of h(x) = 1

x2 + 3 at points wehere x > 0, determine the

equation of the one having the least slope.

Answer: The objective function is:

h(x) = 2x(x2 + 3)2

There will be two candidates, x =1 x = 1. Of them, x =1 will succeed the secondderivative test. Evaluating, h(1) = 1

8. Use this result to find the equation, via the point-

slope form, at the point (1, h(1)) of course. In the end, you should arrive at y= x 38

2 Selected Problems

Some of these problems may challenge you :)

1. Prove that the equationx3 + 2x+ 5 = 0 cannot have more than one real root.

Answer:

Page 10 of 14


11/14



Proof. Let f(x) = x3 + 2x+ 5. To prove the existence of at least one real root, we consider

f(0) = 5 and f(2) =7. By the Intermediate Value Theorem, there is at least one root inthat interval. We have then shown that we have at least one root.

Now assumea, b to be two real roots. Since our function is continuous and differentiable every-where, by Rolles Theorem, we have:

f(a) f(b)a b = 0 =f

(c) = 3x2 + 2

Obviously, we do not have a solution for c in the real numbers which contradicts the result of

Rolles Theorem. So it must be the case that we cannot have both f(a) and f(b) be equal to

zero. Therefore f(x) cannot have more than one real root.

2. Show that the equation 2x

1 = sin x has exactly one real root.

Answer:

Proof. Let g(x) = 2x 1 sin x. To prove we have at least one root, considerg(0) =1 andg() = 21> 0. By the Intermediate Value Theorem, there is at least one root in the interval.

We can proceed with Rolles Theorem, but we have already done that previously. Lets use the

Mean Value Theorem this time.

Leta, bbe two distinct real numbers. Letbbe a root ofg. Sinceg is continuous and differentiable

everywhere, by the Mean Value Theorem we have

g(a) g(b)a b =g

(c)

Sinceg(b) = 0, our equation becomesg(a) =g(c)(ab) = (2cos c)(ab). For the first factor,we are guaranteed that 1 (2 cos c) 3 for any c between a and b.

For the second factor, ifa > b, then a b > 0, and it follows that g (a)> 0. On the other handifa < b, then a b < 0, and g(a)< 0. In both cases, g (a) is non-zero. So g attains it only rootatb as assumed.

Therefore we have shown that g (x) has exactly one real root.

3. Supposef(x) is increasing on (0, 1]. Show that f

1

x2 + 1

is decreasing on (0, 1].

Answer:

Proof. Let f be increasing as given. Then f(0) < f(1) or 0 < f(1) f(0). Dividing by the

Page 11 of 14


12/14



non-negative real number 1 0, we have

0 0 for some c (0, 1).

Let this c be some function ofx, c(x) = 1x2+1

and g (x) =f(c(x)) =f

1

x2 + 1

. By the chain

rule,

g(x) =f(c(x)) c(x)

=f(c(x))

2x

(x2

+ 1)2

At this point, its convenient to suppress c(x) to just c. Doing so, and after rearrangement,

g(x)2x

(x2+1)

=f(c)> 0

Recall that 0< c= 1x2+1

0 org(x)< 0 on this interval.

By the increasing function theorem, g (x) =f 1x2+1 must be decreasing on the interval.

4. An automobile travels 4 km along a straight road in 5 min. Show that the speedometer reads exactly

48 km/h at least once during trip.

Answer:

Proof. Let the distance travelled, x, be a function of time t and let x(0) = 0

x(5) = 4. Since

this is Math and not Physics, we can assume that measured distance and time are continuous

objects and that x(t) is differentiable on t (0, ). By the MVT we havex(5) x(0)

5 0 =4 0

5 =

4

5 =x(c) (1)

Where c (0, 5). A speedometer measures the instantaneous rate of change of the travelleddistance with respect to time so it is equal to x(t) for all t in the domain. Proving that 4

5

km/min is equivalent to 49 km/h is left as an exercise.

5. Use the Mean Value Theorem to prove that if 0 < x < y, theny x < y

x

2x . In particular,show that the geometric mean ofx and y is less than their arithmetic mean, i.e.

xy

1

c >

1

y (2)

Recall that if 0 < a < b and g(x) is strictly increasing then g(a) < g(b). We use this on the

inequation (2) to get

g

1

x

> g

1

c

1

x>

1c

(3)

Multiplying (3) by a positive constant, 1

2

1

2

x>

1

2

c (4)

Equation (1) and the inequation (4) implies

y xy x

= 1

2c

math 53 le 3 reviewer problems

Documents