Math 103. First Exam. Solutions

(a)




lim x® 1  Öx-1 x+2


(b)




lim x® 3  x2-3x x2+x-12


(c)




lim x® 1  1-Öx 1-x

Solution. (a) By the properties of limits

lim x® 1  x1/2-1 x+2 = lim x® 1  (x1/2-1) lim x® 1  (x+2) = 1-1 1+2 = 0 3 =0

(b) This is an indeterminate form 0/0. Thus we factor

lim x® 3  x2-3x x2+x-12 = lim x® 3  x(x-3) (x+4)(x-3) = lim x® 3  x (x+4) = 3 7

(c) This is an indeterminate form 0/0. In this case we rationalize.

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

2. (a) Write the definition of derivative f¢(x) of a function f(x).

(b) Use the definition of derivative in part (a) to find an equation of the line tangent to the graph of the function.

f(x)= 1 x-1

at the point (-1,-1/2).

Solution. (a) The derivative f¢(x) of f(x) is

f¢(x) = lim h® 0  f(x+h)-f(x) h

(b) The slope of the line is the derivative f¢(-1).

f¢(-1) = lim h® 0  1 -1+h-1 - 1 -1-1 h = lim h® 0  -1-1-(-1+h-1) h(1-1)(-1+h-1) = lim h® 0  -1 (-2)(h-2) = -1 4

Thus the equation y=(-1/4) x+b, and plugging in (-1,-1/2) yields -1/2=1/4 +b, thus b=-3/4. An equation of the line is

y = -1 4 x - 3 4

3. The demand and supply equations for a certain commodity are p=-2x²+80 and p=15x+30, respectively, where x represents quantity in units of a thousand and p is the unit price in millions of dollars. Find the equilibrium quantity and equilibrium price.

Solution. The equilibrium equation is

-2x2+80=15x+30

or

2x2 +15x -50=0

Solving for x using the quadratic formula (discard the negative value) yields x=2.5, and so p=67.5. Thus EQ = 2,500 units and EP = $67.5 millions.

4. The number of housing starts per year is estimated to be

N(r) = 1 1+r2

million units, where r (percent) is the mortgage rate. The mortgage rate t months from now is estimated to be

r(t) = t+1 t+2        percent/year

(a) Find an expression for the number of housing starts per year N as a function of t, and use it to determine the number of housing starts 10 months from now.

(b) Use the expression for N(t) found in part (a) to find the number of housing starts in the long run. (Hint: Simplify N(t) and find lim_t®¥N(t).)

Solution. (a) Plug the formula for r(t) into N:

N(t)=N(r(t)) = 1 1+ æ ç è t+1 t+2 ö ÷ ø 2

After simplification you get

N(t)= t2+4t+4 2t2 +6t+5

When t=10, N(10) = 155/265, or 584,905 housing starts.

(b) The highest power of t in both denominator and numerator of N(t) is t². Thus

lim t® ¥  t2+4t+4 2t2 +6t+5 = lim t® ¥  1 t2 (t2+4t+4) 1 t2 (2t2 +6t+5) = lim t® ¥  1+ 4 t + 4 t2 2+ 6 t + 6 t2 = 1+0+0 2+0+0 = 1 2

of 500,000 housing starts.

5. The market value of certain painting own by the reputed art dealer Mr. Greedy has been estimated to grow linearly with its age.

(a) If the value of the painting was $15,400 in 1973 and $41,000 in 1989, what was its value last year?

(b) If Mr. Greedy paid $3,400 for the painting, on what year should he sell it if he wants to make a profit of $60,000?

Solution. (a) The painting increases by

41000-15400 1989-1973 = 25600 16 =1600

dollars per yer. Thus in 2001 its price is

1600(2001-1989)+41000 = 60200

(b) To make a profit of 60,000, he must sell it by 63,400. This would be in 2003.

6. Let f(x) be the function

f(x) = ì ï í ï î x2 if x < 1 2 if x=1 3-2x if x > 1

(a) Sketch the graph of f.

(b) Find the limit lim_{x® 1} f(x), if it exists.

(c) Is f continuous at x=1? Explain.

(d) Is f(x) differentiable at x=1? Explain.

Solution. (a)

(b) The right limit at x=1

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

and the left limit at x=1

lim x® 1-  f(x) = lim x® 1  x2 = (1)2=1

Thus the limit at x=1 exists because righ and left limits are equal, and

lim x® 1  f(x) = 1

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

lim x® 1  f(x) = 1 ¹ f(1)=2

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

7. 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=4 seconds.

(b) What is the average velocity of the object over the time interval [4,5]?

(c) Find the maximum altitude attained by the object. (Hint: At its highest point, its velocity is zero.)

Solution. The velocity at time t is the derivative f¢(t). Thus

f¢(t)= lim h® 0  100(t+h)-(t+h)2 -100t-t2 h = lim h® 0  100 t +100 h -t2 -h2 -2th -100 t -h2 h = lim h® 0  100 h -h2-2th h = lim h® 0  (100 -h -2t) = 100-2t

(a) f¢(4) = 100-2(4)=92

(b) The average velocity

distance covered time elapsed = f(5)-f(4) 5-4 = 100(5)-(5)2-100(4)+(4)2 1 = 91

(c) At highest point velocity is 0. Thus 100-2t=0, so t=50. The height at the moment when t=50 is f(50)=100(50)-(50)² = 5000-2500=2500 ft.

8. Juanita wishes to have a rectangular garden in her backyard. She wants her garden to have an area of 250 ft². Letting x denote the width of the garden, find a function f(x) in the variable x giving the length of the fencing material required to construct the garden. What is the domain of the function f(x)?

Solution. If x denotes the width and y denotes the length of the garden, then the area is xy=250, thus y=250/x.

The perimeter is f=2x+2y. Substituting y for y=250/x in the perimeter it obtains that

f(x)=2x+ 500 x

The domain of this function is (0,¥), that is, all positive numbers.