MATH 180: ELEMENTS OF CALCULUS I
Exam #2 – Solutions
W. Stalin Rios
1
Concept Questions
1. Explain each term in your own words:
(a) Marginal cost function is the derivative of the cost function. It represents how the
cost of a certain product or unit changes over time.
(b) Average cost function is the ratio of the total production cost to the number of units
produced.
(c) Marginal revenue function is the derivative of the revenue function. It measures the
rate of change of the revenue realized from the sale of a certain product.
2. When finding critical points, why do we set the first derivative of f (x) to zero?
Because critical points have a horizontal tangent line which has a slope equal to zero.
2
Differentials
3. Find the differential dy of the function:
√
f (x) = 4x 2x − 4
First, using the product rule we obtain
√
f 0 (x) = 4 2x − 4 +
√
(4x)(2)
4x
√
= 4 2x − 4 + √
(2) 2x − 4
2x − 4
Then, since dy = f 0 (x)dx it yields
dy =
√
4x
4 2x − 4 + √
2x − 4
dx
Marginal Functions in Economics
4. The management of Acrosonic plans to market the ElectroStat, an electrostatic speaker
system. The marketing department has determined that the demand for these speakers is
p = −0.04x + 800
1
(0 ≤ x ≤ 20, 000)
where p denotes the speaker’s unit price (in dollars) and x denotes the quantity demanded.
Acrosonic’s production department estimates that the total cost (in dollars) incurred in
manufacturing x ElectroStat speaker systems in the first year of production will be
C(x) = 200x + 300, 000
Find the marginal profit function.
First, we find the revenue function
R(x) = px = (−0.04x + 800)x = −0.04x2 + 800x
The, the profit function
P (x) = R(x) − C(x)
= −0.04x2 + 800x − (200x + 300, 000)
= −0.04x2 + 800x − 200x − 300, 000
= −0.04x2 + 600x − 300, 000
Finally, the marginal profit function P 0 (x) is given by
P 0 (x) = −0.08x + 600
where P 0 (x) is measured in dollars per number of speakers. We can conclude that the
profit is increasing from x = 0 to x = 7500 speakers, since P 0 (0) = 600 > 0. After that, the
profit is going to start decreasing, since P 0 (7501) = −0.08 < 0. Hence, x = 7500 speakers
(relative maximum) is the number of speakers needed to maximize the profit.
3
Higher–Order Derivatives
GDP of a Developing Country
5. A developing country’s gross domestic product (GDP) from 2000 to 2008 is approximated
by the function
G(t) = −0.2t3 + 2.4t2 + 60 (0 ≤ t ≤ 8)
where G(t) is measured in billions of dollars, with t = 0 corresponding to 2000.
(a) Compute G0 (0)
First, G0 (t) = −0.6t2 + 4.8t, then G0 (0) = 0.
(b) Compute G00 (0)
Since, G00 (t) = −1.2t + 4.8, then G00 (0) = 48.
(c) Interpret your results from parts (a) and (b)
In 2000, since G0 (0) = 0 and G00 (0) = 48 > 0, we conclude that the GDP had a
relative minimum of G(0) = 60 billions of dollars, after which it started increasing.
2
4
Implicit Differentiation
6. Find the equation of the tangent line to the graph of
x2 y + xy 2 − y = 2x + 2
at the point where x = 1.
Hint: use implicit differentiation to find the slope of the tangent line.
First, substituting x = 1 into the given equation we obtain y + y 2 − y = 2 + 2 or y 2 = 4,
hence y = ±2. Then, differentiating implicitly both sides of the given equation yields
2xy + x2
dy
dy
dy
+ y 2 + 2y
−
=2
dx
dx dx
Evaluating
dy
2 − 2xy − y 2
= 2
dx
x + 2y − 1
⇒
dy
dy
3
at the point (1, 2) results
=− .
dx
dx
2
Finally, solving for the y–int at the point (1, 2), we find 2 = −3/2(1) + b ⇒ b = 7/2.
Therefore, the equation of the tangent line at (1, 2) is
3
7
y =− x+
2
2
5
Applications of the Derivative
7. Find the relative maxima and relative minima, if any, of each function. Also, determine
the intervals where each function is increasing and the intervals where it is decreasing.
Hint: use either the first derivative test or the second derivative test.
(a) f (x) =
2x
x2 + 1
First, using the quotient rule we find the first derivative of f (x)
f 0 (x) =
2(x2 + 1) − 2x(2x)
−2x2 + 2
= 2
2
2
(x + 1)
(x + 1)2
Then, to find critical points we solve f 0 (x) = 0, thus −2x2 + 2 = 0
⇒
x = ±1
Now, evaluating test points on both sides of the critical points we obtain
f 0 (−2) = −6/25 ,
f 0 (0) = 2 ⊕,
and f 0 (2) = −6/25 Therefore, f (x) has a relative minimum at x = −1, and a relative maximum at x = 1.
Also, f (x) is increasing on the interval (−1, 1) and it is decreasing on the interval
(−∞, −1) ∪ (1, ∞).
3
(b) g(x) =
1
(x + 3)2
We can first rewrite g(x) = (x + 3)−2 then the first derivative is
g 0 (x) = −2(x + 3)−3 =
−2
(x + 3)3
Then, to find critical points we solve g 0 (x) = 0, thus −2 6= 0. Therefore, there aren’t
any critical points.
Since g 0 (x) is not defined at x = −3 we evaluate the sign of g 0 (x) on both sides of
x = −3. We then obtain
g 0 (−4) = 2 ⊕,
and g 0 (−2) = −2 Therefore, g(x) is increasing on the interval (−∞, −3) and it is decreasing on the
interval (−3, ∞).
8. Sketch the graph of the function, using the curve-sketching guide covered in class.
f (x) = 2.25 + 3x2 +
x4
x3
−
3
4
Hint: the x-intercepts are x1 = −3 and x2 = 4.26
1 Domain: x ∈ R
2 Intercepts: (x = 0) then y–int = 2.25
3 End behavior: f (x) ≈ −x4 hence as x → ∞, y → −∞ and as x → −∞, y → −∞
4 No vertical or horizontal asymptotes since f (x) is a polynomial.
5 f 0 (x) = 6x + x2 − x3 = −x(x2 − x − 6) = −x(x − 3)(x + 2)
Then solving f 0 (x) = 0 yields critical points: x = −2, x = 0, and x = 3. Evaluating
f 0 (x) at some test points we obtain: f 0 (−3) = ⊕, f 0 (−1) = , f 0 (1) = ⊕, and
f 0 (4) = . Thus, f (x) is increasing on the interval (−∞, −2) ∪ (0, 3), and f (x) is
decreasing on the interval (−2, 0) ∪ (3, ∞)
6 Based on the first derivative test, f (x) has relative maximums at x = −2 and at x = 3,
and f (x) has a relative minimum at x = 0.
7 f 00 (x) = 6 + 2x − 3x2
Using the√quadratic formula to solve f 00 (x)
√ = 0, we find inflection points located at
x = (1 − 19)/3 ≈ −1.1 and at x = (1 + 19)/3 ≈ 1.8
Evaluating f 00 (x) at some test points we obtain: f 00 (−10) = , f 00 (0) = ⊕, and
f 00 (10) = . Therefore, f (x) is concave downward on the interval (−∞, −1.1) ∪
(1.8, ∞), and f (x) is concave upward on the interval (−1.1, 1.8)
4
f (x)
Rel. Max
20
15
Infl. Pt.
Rel. Max
10
Infl. Pt.
5
Rel. Min
−3
−2
−1.1
x
1.8
−5
−10
−15
−20
5
3
4.26