Exam 1 Solutions

Math 135 Business Calculus Exam 1 Solutions 1. (20 points) The graph of a function f is shown. (a) Determine the value of each of the following limits, if it exists. If it does not exist, explain why not. (i) lim f (x)

Spring 2009 y

5 4

x→−2

3

As x approaches −2 from either side, the function values get closer and closer to 2, so

2 1

lim f (x) = 2.

x→−2

(ii) lim f (x)

–5

x→−1

As x approaches −1 from the left, the function values get unboundedly large, so the left-hand limit is lim − f (x) = ∞. x→−1

As x approaches −1 from the right, the function values get unboundedly large negative, so the righthand limit is

–4

–3

–2

–1

1

2

3

4

–1 –2 –3 –4 –5

lim f (x) = −∞.

x→−1+

It follows that the two-sided limit lim f (x) does x→−1

not exist. (iii) lim− f (x) and lim+ f (x) x→2

x→2

As x approaches 2 from the left, the function values approach 2, so the left-hand limit is lim f (x) = 2.

x→2−

As x approaches 2 from the right, the function values approach 3, so the right-hand limit is lim f (x) = 3.

x→2+

(iv) lim f (x) x→2

Since the one-sided limits lim− f (x) and lim+ f (x) are not equal, then the two-sided limit x→2

x→2

lim f (x) does not exist.

x→2

(b) Determine whether or not the function is continuous at each of the following numbers. If it is continuous, explain why it is using the definition. If is not continuous, explain why it is not using the definition. (i) a = −2 For a = −2, the function value is f (−2) = 1 while the limit as x approaches −2 is lim f (x) = 1. Since x→−2

lim f (x) 6= f (−2),

x→−2

then f is not continuous at −2. (ii) a = 2 Since the limit lim f (x) does not exist, then f is not continuous at 2. x→2

5

x

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Math 150A Exam 1 Solutions

2. (10 points) Find each limit, if it exists, and indicate how your are computing the limit. If a limit does not exist, state why not. (a) lim (2x3 − x2 + 7x) x→1

Since the function 2x3 − x2 + 7x is a polynonial, then the limit can be computed by evaluating the function at x = 1, so lim (2x3 − x2 + 7x) = 2(1)3 − (1)2 + 7(1) = 8.

x→1

Alternatively, using the Limit Properties, we have lim (2x3 − x2 + 7x) = lim (2x3 ) − lim x2 + lim 7x x→1 x→1 x→1 ≥ ¥3 ≥ ¥2 = 2 lim x − lim x + 7 lim x

x→1

x→1 3

x→1

x→1

2

= 2(1) − (1 ) + 7(1) = 8

(b)

x2 + 2x − 35 x→−7 x+7 Since the limit of the denominator is 0, we cannot apply the Limit Properties. Instead, we can simplify the function algebraically first: lim

x2 + 2x − 35 (x + 7)(x − 5) = lim = lim (x − 5) x→−7 x→−7 x→−7 x+7 x+7 Now, since x − 5 is a polynomial, we can evaluate the limit by direct substitution: lim

x2 + 2x − 35 = lim (x − 5) = −7 − 5 = −12. x→−7 x→−7 x+7 lim

3. (8 points) Based on data from Major League Baseball, the average price of a ticket to a major league baseball game can be approximated by p(x) = 0.03x2 + 0.56x + 8.63, where x is the number of years after 1991 and p(x) is in dollars. (a) Find p(14) − p(4) and interpret this result. To evaluate p(4) and p(14), substitute t = 4 and t = 14 into the function: p(4) = 0.03(42 ) + 0.56(4) + 8.63 = 11.35 p(14) = 0.03(142 ) + 0.56(14) + 8.63 = 22.35 p(4) is the average price of a ticket to a baseball game at t = 4, in 1995, and p(14) is the average price of a ticket at t = 14, in 2005. Then p(14) − p(4) is the increase in the average price of a ticket form 1995 to 2005. p(14) − p(4) (b) Find and interpret this result. 14 − 4

p(14) − p(4) 22.35 − 11.35 11 = = = 1.10 14 − 4 14 − 4 10 This is the average rate of change per year in the average price of a ticket to a baseball game from 1995 to 2005. It tells us that the ticket price increased by $1.10 per year, on average, over this time interval.

Math 150A Exam 1 Solutions

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4. (12 points) Given the function f (x) = 5x2 − 2x + 1, find f 0 (x) using the definition of the derivative. f (x + h) − f (x) h→0 h [5(x + h)2 − 2(x + h) + 1)] − [5x2 − 2x + 1] = lim h→0 h [5(x2 + 2xh + h2 ) − 2x − 2h + 1] − 5x2 + 2x − 1 = lim h→0 h 2 10xh + 5h − 2h h(10x + 5h − 2) = lim = lim h→0 h→0 h h = lim (10x + 5h − 2) = lim 10x + lim 5h − lim 2

f 0 (x) = lim

h→0

h→0

h→0

h→0

= 10x + 0 − 2 = 10x − 2 5. (10 points) Find the equation of the tangent line to the graph of f (x) = 3x − x2 at x = 1. The graph of f (x) = 3x − x2 is shown in the figure. Graph this tangent line. First, find the derivative f 0 (x):

y

d (3x − x2 ) = 3 − 2x dx The slope of the tangent line to the graph at x = 1 is the derivative at x = 1. So

4

mtangent = f 0 (1) = 3 − 2(1) = 1.

2

f 0 (x) =

When x = 1, then y = f (1) = 3(12 ) − 1 = 2. So the tangent line passes through the point (1, 2). Then, substituting into the point-slope formula for a line, the equation of the tangent line is y − y0 = m(x − x0 ) y − 2 = 1(x − 1) y =x+1

3

1

–2

1

–1 –1

–2

2

3

4

x

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Math 150A Exam 1 Solutions

6. (24 points) Find the derivatives of the following functions. Show each of your steps in calculating these derivatives and indicate which differentiation rules are being applied. √ (a) y = (2x5 + 1) x + 1 √ First, apply the Product Rule. To differentiate the second factor x + 1 = (x + 1)1/2 , we need to apply the Extended Power Rule. µ ∂ √ dy d d = (2x5 + 1) · x + 1 + (2x5 + 1) · (x + 1)1/2 dx dx dx ∑ ∏ √ 4 5 −1/2 d 1 = (10x ) x + 1 + (2x + 1) 2 (x + 1) (x + 1) dx √ = 10x4 x + 1 + (2x5 + 1)( 12 )(x + 1)−1/2 (1) √ = 10x4 x + 1 + 12 (2x5 + 1)(x + 1)−1/2 x3 + 2 2x2 + 1 Apply the Quotient Rule. µ

(b) h(x) =

∂ d 3 d (x + 2) · (2x2 + 1) − (x3 + 2) · (2x2 + 1) dx dx h0 (x) = (2x2 + 1)2 (3x2 )(2x2 + 1) − (x3 + 2)(4x) = (2x2 + 1)2 4 2 6x + 3x − 4x4 − 8x 2x4 + 3x2 − 8x = = (2x2 + 1)2 (2x2 + 1)2

(c) y = (x3 + x2 + x)7 Apply the Extended Power Rule. dy d 3 = 7(x3 + x2 + x)6 · (x + x2 + x) dx dx = 7(x3 + x2 + x)6 (3x2 + 2x + 1) (d) y = (5 − x)3 (3x − 1)4 First, apply the Product Rule. To differentiate each of the factors, apply the Extended Power Rule. µ ∂ µ ∂ dy d d 3 4 3 4 = (5 − x) · (3x − 1) + (5 − x) · (3x − 1) dx dx dx µ ∂ µ ∂ 2 d 4 3 3 d = 3(5 − x) (5 − x) · (3x − 1) + (5 − x) · 4(3x − 1) (3x − 1) dx dx = 3(5 − x)2 (−1)(3x − 1)4 + (5 − x)3 (4)(3x − 1)3 (3) = −3(5 − x)2 (3x − 1)4 + 12(5 − x)3 (3x − 1)3

Math 150A Exam 1 Solutions

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7. (16 points) Given the function f (x) = 4 + 3x − x3 . (a) Find the critical values. First, find the derivative f 0 (x). d (4 + 3x − x3 ) = 3 − 3x2 dx Second, set the derivative equal to 0 and solve for x. f 0 (x) =

f 0 (x) = 3(1 − x2 ) = 0 3(1 − x)(1 + x) = 0 x=1 or x = −1

The critical values are x = −1 and x = 1. (b) Use f 0 (x) to determine where the function is increasing or decreasing and to find the relative maxima and minima. Plot the critical values on a number line. This divides the number line into three intervals. Choose a test value in each interval and determine whether f 0 (x) is positive or negative at each test value. f incr/decr

decreasing –2

f' +/–

–

–

–

increasing 0 –1

–

+

+

decreasing 2 1

+ +

–

–

–

–

° ¢° ¢ f 0 (−2) = 3 1 − (−2) 1 + (−2) < 0 f 0 (0) = 3(1 − 0)(1 + 0) > 0 f 0 (2) = 3(1 − 2)(1 + 2) < 0

This tells us that f is decreasing for x < −1, increasing for −1 < x < 1, and decreasing for x > 1. From the First-Derivative Test, it follows that f (−1) is a relative minimum and f (1) is a relative maximum. (c) Sketch a graph of the function. Indicate the relative maxima and minima on the graph. y 6

x

y = f (x)

−4

56

−3

22

−2

6

−1

2

0

4

1

6

2

2

3

−14

–3

4

−48

–4

relative maximum

5 4 3

relative minimum

2 1

–6

–5

–4

–3

–2

–1

1 –1 –2

–5 –6

2

3

4

5

6

x