[PDF] Math 180,Exam 2, Spring 2011 Problem 1Solution

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Math 180,Exam 2, Spring 2011 Problem 1Solution

Math 180,Exam 2, Spring 2011 Problem 2Solution 2 Sketch the graph of the function f(x) = x−2 −x2 following the steps below (a) Determine the domain of f and find all asymptotes (b) Find the intervals where the graph of f is increasing, decreasing, concave up and concave down

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Let the common ratio between a and b be x, ∴a = 2x and b = 3x XY is a straight line, OM and OP rays stands on it Therefore, XOM + MOP + POY = 180 o ⇒b + a + POY + 180o ⇒3x + 2x + 90o= 180o ⇒5x = 90o ⇒x = 18o a = 2x ⇒= 2 × 18 ⇒= 36o b = 3x ⇒= 3 × 18 ⇒= 54o Now, MN is a straight line OX ray stands on it ∠b + ∠c = 180o

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Math 180, Exam 2, Spring 2011

Problem 1 Solution

1. The graph of a functionf(x) is shown below:A B C D E F G

cde (a) Fill in the table below with the signs of the rst and second derivatives offon each of the intervalsA;:::;G.ABCDEFG sign off0sign off00 (b) Which of the pointsa;:::;gare critical points? For each critical point, say whether it is a local maximum, a local minimum or neither. (c) Which of the pointsa;:::;gare in ection points?Solution: (a)

ABCDEFG

sign off0+++++ sign off00+++ (b)c,e, andgare critical points becausef0(x) = 0 at these points.cis neither a local minimum nor a local maximum.eis a local maximum.gis a local minimum. (c)c,d, andfare in ection points becausef00(x) changes sign at these points.

Math 180, Exam 2, Spring 2011

Problem 2 Solution

2. Sketch the graph of the functionf(x) =x-2x2following the steps below.

(a) Determine the domain offand nd all asymptotes. (b) Find the intervals where the graph offis increasing, decreasing, concave up and concave down. (c) Sketch the graph off, clearly showing any local extrema, in ection points,x-intercepts, y-intercepts and asymptotes.

Solution:

(a) The domain offis all real numbers exceptx= 0. In fact,x= 0 is a vertical asymptote.

Furthermore, since

limx!1-x-2x2= we know thatfdoes not have a horizontal asymptote. (b) To determine the intervals of monotonicity, we rst nd the critical points off. These are points where eitherf0(x) = 0 orf0(x) does not exist. The rst derivative is:

0(x) =2x-32x=2x2

0(x) will not exist only whenx= 0. However,x= 0 is not in the domain off.

Therefore, the only critical points we have are solutions tof0(x) = 0.

0(x) = 0

2x2 x3= 0 x+1 = 0

2x4+ 1

= 0

4+ 1 = 0

This equation has no solutions. Thus,fhas no critical points. To determine the intervals of monotonicity we take the domain offand evaluatef0(x) at test points to determine the sign off0(x).

IntervalTest Point,cf0(c)Sign off0(c)

(;0)1f0(1) = 4+ (0;)1f0(1) =4 From the table we determine thatfis decreasing on (0;) becausef0(x)<0 for all x(0;) andfis increasing on (;0) becausef0(x)>0 for allx(;0). To determine the intervals of concavity, we rst nd the values ofxfor whichf00(x) = 0.

00(x) = 0-2x2x-30= 0

2 + 6x-4= 0

2 +6 x4= 0

2x4+ 6 = 0

4= 3 x=4 To determine the intervals of monotonicity we split the domain offinto the intervals (;4

3), (43;0), (0;43), and (43;) and evaluatef00(x) at test points in each

interval to determine the sign off00(x).

IntervalTest Point,cf00(c)Sign off00(c)

(;43)2f00(2) =138 (43;0)1f00(1) = 4+ (0;43)1f00(1) = 4+ (43;)2f00(2) =138 From the table we determine thatfis concave down on (;43)(43;) because

00(x)<0 for allx(;4

3)(43;) andfis concave up on (43;0)(0;43)

becausef00(x)>0 for allx(4

3;0)(0;43).

43114323xy

Math 180, Exam 2, Spring 2011

Problem 3 Solution

3. Find the area of the largest rectangle that can be inscribed in a semicircle of radius 3.

Figure 1:

Solution: Letxbe half of the width of the rectangle and letybe the height. We seek to maximize the area so the function is:

Area = 2xy

The variablesxandycan be related using the Pythagorean Theorem.

2+y2= 32

Solving the above equation forywe get:

y=⎷ 9-x2

Plugging this into the area formula we get:

Area = 2xy

f(x) = 2x⎷ 9-x2 We must now nd the absolute maximum off(x) on the domain [0,3]. We start by ndingquotesdbs_dbs7.pdfusesText_5