M403K First Midterm Exam Solutions.  September 21, 2000

1. Consider the function tex2html_wrap_inline62

a) Find f'(x).

Soln: tex2html_wrap_inline66

b) Find the slope of the line tangent to the curve y=f(x) at x=0.

Soln: Slope equals tex2html_wrap_inline72

c) Find the equation of this tangent line.

Soln: Since f(0)=1, the tangent line is y-1 = 2.4(x-0), or y = 2.4 x + 1.

d) Use this tangent line to approximate f(0.01).

Soln: So f(0.01) is very close to 2.4(0.01) + 1 = 1.024.

2. From the following table, estimate f'(1). Indicate clearly how you obtain your answer:

tabular8

Solution: For h ranging from -.03 to +0.03, look at [f(1+h)-f(1)]/h:

tabular11

As h approaches zero from either side, [f(1+h)-f(1)]/h approaches 3, so f'(1) equals 3.

3. Evaluate the following limits, if they exist (or write DNE if they do not).

a) tex2html_wrap_inline108

As tex2html_wrap_inline110 , the numerator goes to zero while the denominator goes to -3, so the ratio goes to zero.

b) tex2html_wrap_inline114

As tex2html_wrap_inline116 both the numerator and denominator go to zero, so we have to be more careful. The numerator equals (x-2)(x+1). Canceling a factor of (x-2) from numerator and denominator we have

displaymath122

c) tex2html_wrap_inline124

As tex2html_wrap_inline126 , the numerator goes to 1 while the denominator goes to zero from above, so the limit is tex2html_wrap_inline128 . Put another way, if x is slightly bigger than 1 (say 1.001), then the numerator is close to 1 while the denominator is a small possitive number (like .001), so the ratio is a large positive number (like 1000).

d) tex2html_wrap_inline132

As tex2html_wrap_inline134 , the numerator goes to 6, but the denominator goes to zero. If x is slightly bigger than 4, then the denominator is a small positive number and the ratio is a large positive number. That is, tex2html_wrap_inline138 . But if x is slightly less than 4, then the denominator is a small negative number and the ratio is a large negative number. That is, tex2html_wrap_inline142 . Since the limits from the two sides do not agree, there is no overall limit. The correct answer is ``DNE''.

4. Take the derivatives of the following functions with respect to x. You do not need to simplify your answers:

a) tex2html_wrap_inline146

Use the chain rule: tex2html_wrap_inline148 .

b) tex2html_wrap_inline150

Use the product rule: tex2html_wrap_inline152 .

c) tex2html_wrap_inline154

You can either use the quotient rule or the chain rule. Since tex2html_wrap_inline156tex2html_wrap_inline158 .

d) tex2html_wrap_inline160

Use the quotient rule. tex2html_wrap_inline162 .

5. New England Widget Technologies (NEWT) makes expensive high-tech widgets. Their marketing department has determined that the demand function is x = 4000 - 2p, where x is the number of widgets sold and p is the price. Their cost function is C(x)= 200,000 + 1000x.

a) Find the price p(x) and the revenue R(x) as a function of x.

Soln: p(x) = (4000-x)/2 = 2000 - x/2. tex2html_wrap_inline180 .

b) Compute the marginal cost, marginal revenue and marginal profit as a function of x.

Soln: Marginal cost = C'(x) = 1000. Marginal revenue = R'(x)=2000-x. Marginal profit = P'(x) = R'(x)-C'(x) = 1000 - x.

c) The company has a current production level of x=1500. To increase revenue, should the company increase or decrease production? [Note: you do not need to compute the optimal level of production. You just need to say whether it is higher or lower than 1500.]

Soln: Since R'(1500)=2000-1500=500, each additional widget produced will increase revenue by $500. So to increase revenue the company should increase production.

c) The company has a current production level of x=1500. To increase profit, should the company increase or decrease production?

Soln: Since P'(1500)=1000-1500=-500, each additional widget made will decrease profits by $500. This makes sense, because each additional widget costs $1000 to make and only brings in $500 in revenue. So, to increase profit, the company should decrease production.


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Lorenzo Sadun

Wed Oct 4 11:17:58 CDT 2000