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Spring 2017 Applied Calculus Final Exam Brief Solutions
Howard University Math Department
INSTRUCTIONS:
Each problem 20 points
Total time 2 hours
You must provide step by step solutions.
PARTI:DOALLPROBLEMS
1. Find the equation of the tangent line to the graph of f(x) = (x2+5x−2)2 at the point
where x = 1. You must use chain rule to find the derivative first.
Soln: At x = 1 you get f(1) = y = 16. Slope of tangent at 1 is given by derivative
f′(1) = (12 + 5(1) − 2)(2(2(1) + 5) = 4(14) = 56. Equation is y = 56x − 40.
2. The total cost of producing q units of a certain commodity is given by
C(q) = 3q2 +q +500.
a) What is the total cost to produce 10 units?
b) Use marginal analysis to estimate the cost of producing the 10th unit.
Soln: (a) C(10) = 810. (b) Need C′(9) = 6(9) + 1 = 55 because you want the change
during 10th unit, so at start of that period q = 9.
3. Given that the unit price is p(q) = 20 − q and the total cost is C(q) = 2q2 + 7q + 50
for q units, find the revenue function R(q), profit function P(q), marginal cost C′(q),
average cost A(q), and the level of production where A(q) is minimized.
Solution:
R(q) = qp(q) = q(20−q) = 20q −q2.
P(q) = R(q)−C(q) = −3q2 +13q −50 and marginal profit P′(q) = R′(q)−C′(q) =
37−4q−(6q+5)=32−10q.
Marginal cost C′(q) = 4q +7.
Average cost is A(q) = C(q)/q = 2q +7+ 50.
q
Average cost is minimized when average cost equals marginal cost.
[You can also do this by taking derivative of A(q) and setting it equal to 0, then solving
for q].
So we set A(q) = C′(q) to get 2q + 7 + 50 = 4q + 7.
q
Solving this, we get 2q = 50/q =⇒ q2 = 25 =⇒ q = 5 (q cannot be negative, so we
take only positive root).
1
4. 1000 dollars are deposited in an account at 5 percent annual interest, compounded
continuously. Let A(t) be the amount in the account after t years.
(a) At what rate is A(t) increasing at the end of 3 years?
(b) How long will it take to double (equal 2000) ?
Solution:
4a) Need to find derivative!
Here P = 1000,r = 0.05,t = 3. We get
rt ′ 0.05×3 0.15
A(t) = Pe =⇒ A(3)=1000e (0.05) = 50e =58.09
4b) We want A(t) = 2000. The rest are as in 3a except that we don’t know what t is.
Weget
0.05t 0.05t
2000 = 1000e =⇒ 2=e
=⇒ ln(2) = ln(e0.05t) = (0.05t) =⇒ t = ln2/0.05 = 13.86 years
5. Find the indefinite integral.
Z 2 √
(a) x + x+1dx
Z x
(b) √dx
1+x
Z 2 √ Z 2 √
x + x+1 1 1 x
Soln: (a) x dx = (x+ √ +x)dx= 2 +2 x+lnx+C
x
√ R √
(b) Use substitution u = 1+x. Then du = dx and the integral becomes du/ u=
√ √
2 u+C=2 1+x+C.
PARTII : DO ANY 5 PROBLEMS
6. Find the following limits:
3x+2 x2 +3x−4
(a) lim 3 (b) lim 2 .
x→∞ 4x −1 x→1 x −1
Soln: (a) Divide above and below by x3. Everything goes to 0 as x → ∞.
(b) Factoring we get (x + 4)(x − 1)/(x + 1)(x − 1). We can cancel x − 1 and then let
x→1(notbefore!) and get the limit as (1+4)/(1+1) = 5/2.
7. Use implicit differentiation to find dy/dx at x = 1,y = 0 given that
x3 −xy+4y =1.
2
Soln: Differentiate both sides, using product rule for xy and remembering that y is a
function of x.
Get 3x2 −y −xdy +4dy = 0.
dx dx
2
Solving for dy/dx we get (3x −y)/(x−4).
Letting x = 1,y = 0 get y′(1) = 3/ − 3 = −1.
8. Find the critical points of f(x) = x and classify each critical point as a rela-
x+1
tive maximum, relative minimum or neither. [Note that critical points are where the
derivative is zero or undefined].
Find the absolute maximum and minimum of f(x) in [0,108].
Soln:
First we find the critical points by solving f′(x) = 0 for x. Note that the points where
f′(x) do not exist are also critical points.
Using quotient rule, we find f′(x) as:
f′(x) = (x)′(x + 1) − (x)(x + 1)′ = 1(x + 1) − x = 1 .
(x+1)2 (x+1)2 (x+1)2
The critical points are where either the numerator or the denominator becomes zero.
[When the numerator is zero, but denominator is not zero, you get f′(x) = 0. When
the denominator alone is zero, the derivative doesn’t exist]
Wesee that numerator is always 1, so never zero. So derivative is never zero.
When denominator is zero we get x+1 = 0 or x = −1.
So -1 is only critical point. But function is undefined at -1, so it is neither relative
maximum nor minimum.
So we divide the number line into two intervals (−∞,−1) and (−1,∞) and test points
from these intervals.
Wesee that f′(x) is positive for all values because it is the square of 1/(x + 1).
So f is increasing always.
Its absolute minimum and maximum have to be at the boundary points because there
are no relative extrema. Now f(0) = 0 and f(108) is almost 1. So 0 and 1 are the
absolute minimum and maximum respectively.
9. Evaluate the following integrals:
(a) Z xe2xdx
3
Z 2
(b) xex dx
Soln:
(a) Integrate by parts with f = x and g′ = e2x. Then f′ = 1 and g = e2x/2. Plugging
into R fg′dx = fg −R f′gdx we get the integral as
(xe2x)/2 −e2x/4+C.
(b) Substitute u = x2. Then xdx = du/2. The integral becomes
Z 2
u u x
e du/2 = e /2+C = e /2+C.
10. Find the average value of f over the interval 0 ≤ x ≤ 3 where
f(x) = √ x
x2 +16
Soln: Z h i
1 x 1 √ 3
Average value = √ dx = x2 +16 =1/3
3 x2 +16 3 0
using the substitution u = x2 + 16 and integrating.
11. Compute the improper integral. If the integral does not converge, write “Divergent.”
(a) Z ∞e3−xdx
3
(b) Z ∞ 1 dx
4 xlnx
Soln:
3−x 3 −x
For (a) use the substitution u = 3 − x,du = −dx or write e =e e and take the
e3 outside integral. You get after integrating
3−xN 3−N 3−3 0
lim −e 3 = lim −e +e =e =1.
N→∞ N→∞
For (b) use the substitution u = ln x,du = 1/x to get
Z ∞ 1 dx = lim [ln(ln x)]N = lim ln(ln(N))−ln(ln(4)) = ∞ (Divergent)
xlnx N→∞ 4 N→∞
4
2 3 3 5
12. Let f(x,y) = 2x y −x y
4
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