Practice Midterm

The size of the midterm will be about half the size of this sample exam.

1. Find the indicated limits, if they exist

(a)




lim x® 1  x-1 Öx-1


(b)




lim x® -¥  x2 x3-9

Solution. (a) This is an indeterminate form. Rationalize:

lim x® 1  (x-1)(Öx+1) (Öx-1)(Öx+1) = lim x® 1  (x-1)(Öx+1) x-1 = lim x® 1  (Öx+1)=1+1=2

(b) The highest power of x in both denominator and numerator is x³. Multiply and divide by 1/x³

lim x® -¥  x2 x3-9 = lim x® -¥  1 x3 x2 1 x3 (x3-9) = lim x® -¥  1 x 1- 9 x3 = 0 1-0 = 0

2. A manufacturer has a monthly fixed cost of $40,000 and a production cost of $8 per unit produced. The product sell for $12/unit.

(a) What is the cost function?
(b) What is the revenue function?
(c) What is the profit function?
(d) Compute the profit (loss) corresponding to productions levels of 8,000 and 12,000 units.

Solution. (a) The cost function is

C(x) = 40,000 + 8x

(b) The revenue function is

R(x) = 12 x

(c) The profit function

P(x)=R(x)-C(x) = 12 x- 40,000 - 8x = 4x -40,000

(d) P(8,000) = -8,000 (loss) and P(12,000)=8,000 (profit).

3. Given the demand equation 3x+p-40=0 and the supply equation 2x-p+10=0 where p is the unit price in dollars and x represents the quantity in units of a thousand of certain commodity, determine the equilibrium price and the equilibrium quantity.

Solution. First solve for p both equations: demand p = 40 -3x and supply p=2x +10. Then equate

40-3x = 2x +10

which yields x=30/5=6. Equilibrium quantity is 6,000 units and equilibrium price is 22.

4. The percentage of people over age 65 who have a high school diploma is known to grow linearly. It was 30% in 1970 and 52% in 1990.

(a) What percentage of people over 65 would have a high school diploma in the year 2002?
(b) On what year would you expect that 85% of people over 65 would have a high school diploma?

Solution. (a) The percentage increase each year is

m = 52-30 1990-1970 = 22 20 =1.1

Thus in the year 2002 the percentage of people over 65 with a high school diploma is 52 + (1.1)12=65.2

(b) For the percentage to go from 52% in 1990 to 85%, it needs to increase by 33%, thus it will be 30 years after 1990, that is, in 2020.

5. Let f(x) be the function

f(x)= ì í î |x+1| if x £ 0, x2 if x > 0

(a) Sketch the graph of f.
(b) Does the limit lim_{x® 0} f(x) exist? (Explain.)

Solution (b) The right limit

lim x® 0+  f(x) = lim x® 0+  x2 = 0

and the left limit

lim x® 0-  f(x) = lim x® 0-  |x+1| = 1

Thus, the limit lim_{x® 0} f(x) does not exist

6. Let f(x)=x²-2x+1.

(a) Find the derivative f¢ of f.
(b) Find the point(s) of the graph of f where its tangent line is horizontal. Sketch the graph of f and its tangent line at this point.
(c) What is the rate of change of f at the point found in part (b)

Solution. (a) By the definition of derivative

f¢(x)= lim h® 0  f(x+h)-f(x) h = lim h® 0  (x+h)2 -2(x+h) +1-(x2 -2x +1) h = lim h®0  x2+2xh+h2-2x-2h+1-x2+2x-1 h = lim h® 0  2x +h-2 = 2x-2

(b) The tangent to the graph is horizontal when its slope is 0. The slope is the derivative f¢(x). Thus solve 2x-2=0, hence x=1. The tangent is horizontal at the point (1,0)

(c) The rate of change is 0.

7. Patricia wishes to have a rectangular garden in her backyard. She has 80 ft of fencing material with which to enclose her garden. Letting x denote the width of the garden, find a function f(x) in the variable x giving the area of the garden. What is its domain?

Solution. Let x be the width and y the length of the garden, respectively. The area is xy.

Also, the perimeter of the garden is 80, so 2x+2y=80, and solving for y we have y=40-x. Plugging this into the formula for the area we have

f(x)=x(40-x)

8. In a certain biology lab experiment, the number of bacteria N is related to the temperature of the environment T by the function

N(T)=-2T2 +240 T -5400,              (40 £ T £ 90)

(N(T) is the number of bacteria present when the temperature is T⁰F.)

The temperature t hours after the experiment begins is given by

T(t) = 10t+40,             (0 £ t £ 5).

(a) Express the number of bacteria N as a function of the temperature t.
(b) How many bacteria are present 2 hours after the experiment begins?

Solution.(a) Plug in the formula for temperature T as function of time t into the formula for number of bacteria N as function of temperature T

N(T(t))=-2(10t+40)2 +240(10t+40) -5400 = -200 t2 +800 t +1000

(b) N(2)=-200(2)² +800(2)+1000 = 1800.

9. Let f(x) be the function given by

f(x) = ì ï í ï î x2 if x < 1, 2 if x=1, 2x-1 if x > 1

(a) Sketch the graph of f(x).
(b) Is f(x) continuous at x=1?
(c) Does f(x) have derivative at x=1?

Solution. (b) The right limit at x=1 is

lim x® 1+  f(x) = lim x® 1+  2x-1 = 2(1)-1=1

and the left limit

lim x® 1-  f(x) = lim x® 1-  x2=1

Thus the limit

lim x® 1  f(x)=1

But f(x) is not continuous at x=1 because

lim x® 1  f(x) = 1 ¹ f(1)=2

(c) f(x) has no derivative at x=1 because it is not continuous there.

10. (a) State the definition of the derivative of a function f(x).

(b) Use the definition of the derivative to find the slope of the line tangent to the graph of the function

f(x)=   ___ Öx-1

at the point (a,f(a)).

Solution. (b)

f¢(a)= lim h® 0  f(a+h)-f(a) h = lim h® 0    _____ Öx+h-1   -   ___ Öx-1   h = lim h®0  x+h-1-x+1 h(   _____ Öx+h-1   -   ___ Öx-1   ) = lim h®0  1 h(   _____ Öx+h-1   -   ___ Öx-1   ) = 1 2   ___ Öx-1

11. Determine the values of x for which the following function f(x) is continuous.

f(x) = ì ï ï ï í ï ï ï î x2-1 (x-2)(x+1) if x ¹ -1,2 2/3 if x=-1 3 if x=2

Solution. From the formula, f(x) is continuous at all points x ¹ -1,2. To study, note that we can write

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

when x ¹ -1,2. Thus

lim x® -1  f(x) = lim x® -1  x-1 x-2 = -1-1 -1-2 = 2 3

This equals f(-1)=2/3, so f(x) is continuous at x=-1.

For x=2 the limit

lim x® 2  f(x) = lim x® 2  x-1 x-2 = 1 0

does not exist. Thus f(x) is not continuous at x=2.

12. An object is moving vertically according to the equation s=f(t)=100t-t², 0 £ t £ 100, where t is the time in seconds and s is the height of the object above the ground.

(a) Find the velocity of the object when t=5 sec.
(b) What is the average velocity of the object over the time interval [5,6]?
(c) Find the maximum altitude attained by the object. (Hint: At its highest point, its velocity is zero.)

Solution. (a) The velocity when t=5 is the derivative f¢(5). Using the definition of derivative,

f¢(5)= lim h® 0  f(5+h)-f(5) h = lim h® 0  100(5+h) -(5+h)2 -100(5)+(5)2 h = lim h® 0  90 h-h2 h =90

Alternatively, using the rules for the derivative, we find f¢(t)=100-2t, and thus f¢(5)=100-2(5)=90.

(b) The average velocity over the interval [5,6] is

distance covered Time elapsed = f(6)-f(5) 6-5 = 100(6)-(6)2-100(5)+(5)2 1 = 89

(c) The velocity at time t is 100-2t. (This is like part (a), with t instead of 5.) Thus the velocity is zero when 100-2t=0, that is, when t=50.

13. Find the derivatives of the following functions.

(a) f(x)=(1+Öx)(2x²-3)
(b) g(x) = (x³+2x+1)(2+ 1/x²)

f¢(x) = (1/2)x-1/2(2x2-3) +(1+Öx)(4x)

(b) Use the product and power rules:

g¢(x) = (3x2+2)(2+1/x2)+ (x3+2x+1)(-2/x3)

14. Find the equation of the line through the point (-1,3) and is perpendicular to the line passing through the points (-3,4) and (2,1).

Solution. The line through the points (-3,4) and (2,1) has slope

1-4 2-(-3) = -3 5

and the perpendicular line has slope m=5/3. The line we are looking for has the formula y=(5/3) x+b. To find out b, plug in x=-1 and y=3 in this formula, and get 3=(5/3) (-1)+b, so that b = 14/3. The equation is

y= 5 3 x+ 14 3

15. The base salary of a salesman working on commission is $12,000. For each $50,000 of sales beyond $100,000, he is paid a $1,000 commission. Sketch a graph showing his earnings f(x) as a function of his sales x. Determine the values of x for which the function f(x) is discontinuous.