Math for Econ I-Worksheet 4New York University

1. Compute the derivatives of the following using the power rule:

(a) x6 (b) 2x2 (c) 3x (d) 2xxSolution:

(a) 6x5

(b) 4x3

(c) 32 x3/2

(d) 3x5/2

2. Compute the following:

(a) ddy (Ayb+1) (b) ddA

(1

A2A

)(c) ddp (Ap

+B)

Solution: (a) ddy (Ayb+1) = A(b+ 1)yb.

(b) ddA

(1

A2A

)= 52 A

7/2

(c) ddp (Ap +B) = Ap1

3. For each of the following functions find a function F (x) that has f(x) as its derivative. (Note that you are notasked to find f (x).)

(a) f(x) = x2 (b) f(x) = 2x+ 3 (c) f(x) = xa

Solution: There are many answers.

(a) x3

3

(b) x2 + 3x

(c) xa+1

a+1

4. In the graph of f in figure below, at which of the labelled x-values is

(a) f(x) greatest?

(b) f(x) least?

(c) f (x) greatest?

(d) f (x) least?

!Solution: (a) x3 (b) x4 (c) x5 (d) x3

5. Let f(x) = 1x . Using the definition of the derivative show that f(x) = 1x2 . Using this find the equation of the

tangent line to the curve at x = 1. Solution:

f (x) = limh0

1x+h 1x

h

= limh0

hx(x+h)

h

= limh0

hx(x+ h)h

= limh0

1x(x+ h)

= 1x2

f (1) = 1 and f(1) = 1. So the tangent line has equation y = 1 (x 1) = x+ 2.6. Figure below shows the graph of f , the derivative of a function f . On what interval(s) is the function f

(a) Increasing?

(b) Decreasing?

!

2.3 THE DERIVATIVE FUNCTION 139

Job: chap2-temp Sheet: 71 Page: 139 (January 18, 2012 16 : 44) [ex-3]

SHORT ANSWER:

time

current

2-3w41 43. Figure 2.80 is the graph of f , the derivative of a function f . On what interval(s) is the function f(a) Increasing? (b) Decreasing?

2-3w41fig

x1

x2

x3 x4 x5

f

x

Figure 2.80: Graph of f , not f

ANSWER:(a) The function f is increasing where f is positive, so for x1 < x < x3.(b) The function f is decreasing where f is negative, so for 0 < x < x1 or x3 < x < x5.

SHORT ANSWER:(a) x1 < x < x3(b) 0 < x < x1; x3 < x < x5

2-3w42 44. The derivative of f is the spike function in Figure 2.81. What can you say about the graph of f?

2-3w42figt

f (t)

Figure 2.81

ANSWER:On intervals where f = 0, f is not changing at all, and is therefore constant. On the small interval where f > 0,

f is increasing; at the point where f hits the top of its spike, f is increasing quite sharply. So f should be constant for awhile, have a sudden increase, and then be constant again. A possible graph for f is shown in Figure 2.82.

2-3w42ans

f(t)

t

Figure 2.82: Step function

SHORT ANSWER:SHORT ANSWER INAPPROPRIATE

Solution:

(a) The function f is increasing where f is positive, so for x1 < x < x3.

(b) The function f is decreasing where f is negative, so for 0 < x < x1 or x3 < x < x5.

7. The cost, C = f(w), in dollars of buying a chemical is a function of the weight bought, w, in pounds.

(a) In the statement f(12) = 5, what are the units of the 12? What are the units of the 5? Explain what thisis saying about the cost of buying the chemical.

(b) Do you expect the derivative f to be positive or negative? Why?

(c) In the statement f (12) = 0.4, what are the units of the 12? What are the units of the 0.4? Explain whatthis is saying about the cost of buying the chemical.

Solution:

(a) The 12 represents the weight of the chemical; therefore, its units are pounds. The 5 represents the cost ofthe chemical; therefore, its units are dollars. The statement f(12) = 5 means that when the weight of thechemical is 12 pounds, the cost is 5 dollars.

(b) We expect the derivative to be positive since we expect the cost of the chemical to increase when the weightbought increases.

(c) Again, 12 is the weight of the chemical in pounds. The units of the 0.4 are dollars/pound since it is therate of change of the cost as a function of the weight of the chemical bought. The statement f (12) = 0.4means that the cost is increasing at a rate of 0.4 dollars per pound when the weight is 12 pounds, or thatan additional pound will cost about an extra 40 cents.

8. The supply function for a commodity with price P is given by S(P ) = a+bP , where a and b are positive constants.Find dSdP using the definition of the derivative and interpret its meaning. Sketch a graph of both S(P ) and S

(P ).Make sure you label the axes.

Solution:

S(P ) = limh0

S(P + h) S(P )h

= limh0

a+ b(P + h) a bPh

= limh0

bh

h= b.

9. The demand curve for a product is given by q = 3003p, where p is the price of the product and q is the quantitythat consumers buy at this price.

(a) Write the revenue as a function, R(p), of price.

(b) Find R(10) and interpret your answer in terms of revenue.

(c) For what prices is R(p) positive? For what prices is it negative?

Solution:

(a) We have R(p) = pq = p(300 3p) = 300p 3p2(b) Since R(p) = 300 6p, we have R(10) = 300 6 10 = 240. This means that revenues are increasing at a

rate of $240 per dollar of price increase when the price is $10.

(c) R(p) = 300 6p is positive for p < 50 and negative for p > 50.10. Let f(x) = x3 6x2 15x + 20. Find f (x) and all values of x for which f (x) = 0. Explain the relationship

between these values of x and the graph of f(x). Determine where f (x) is positive ans where it is negative.

Solution: Since f(x) = x3 6x2 15x+ 20, we have f (x) = 3x2 12x 15. To find when f (x) = 0, we solve

3x2 12x 15 = 03(x2 4x 5) = 0

3(x+ 1)(x 5) = 0.

We see that f (x) = 0 at x = 1 and at x = 5. We can see that the graph of f(x) is horizontal at x = 1 andat x = 5, which confirms what we found using the derivative.

11. For the total cost functionTC(y) = y2 + 10y + 25, y > 0

show that

(a) MC is less than AC where AC is falling.

(b) MC = AC at the point where the AC curve is horizontal.

(c) MC exceeds AC where AC is rising.

Solution: MC(y) = 2y+ 10 and AC(y) = y2+10y+25

y = y+ 10 +25y and AC

(y) = 1 25y2 . AC is horizontal whereAC (y) = 0, so 1 25y2 = 0 implies y = 5. At y = 5 MC = 2 5 + 10 = 20 and AC = 5 + 10 + 5 = 20 hence theyare equal this shows part(b) . Now MC < AC implies

2y + 10 < y + 10 +25

y

y < 5

and this is where AC is falling. This shows part (a). Similar work shows part (c).