Math 205
Exam 2, Solutions
Thursday March 14
Your name:
Instructor’s name:
Time the class meets:
Calculators are not permitted. You can use a note card with formulas prepared in advance. No
books are allowed. Use the back pages as sketch paper. For full credit, show your work in detail.
# 1 # 2 # 3 # 4 # 5 # 6 Total
30
10
20
10
10
20
100
1. (5 points for each part) Differentiate:
(a) −2x.
(b) 2x2 − 7.
(d)
√
(c) 10e0.3x
√
3
ex − 1
e x
(f)
.
1+x
(e) x ln(2 + 3x).
Solution.
(−2x)′ = −2;
(2x2 − 7)′ = 2(x2 )′ = 2 · 2x = 4x.
(10e0.3x )′ = 10(0.3e0.3x ) = 3e0.3x .
(We are using the formula (ekx )′ = kekx .)
√
(
)′ 1
1
( ex − 1)′ = (ex − 1)1/2 = (ex − 1) 2 −1 (ex − 1)′ =
2
1
1
1 x
ex
ex
.
(e − 1)− 2 ex = (ex − 1)− 2 = √ x
2
2
2 e −1
(Any of last three answers is fine. We are using (f (x)a )′ = af (x)a−1 f ′ (x).)
(2 + 3x)′
3x
= ln(2 + 3x) +
.
2 + 3x
2 + 3x
We used the product rule and the formula (ln f (x))′ = f ′ (x)/f (x).
(x ln(2 + 3x))′ = (x)′ ln(2 + 3x) + x(ln(2 + 3x))′ = ln(2 + 3x) + x
(
√
3
e x
1+x
)′
(
=
e
√
3
)
x ′
(1 + x) − e
(1 + x)2
√
3x
(1 + x)′
=
e
√
3x
(x1/3 )′ (1 + x) − e
(1 + x)2
√
3x
=
√
1 3 x −2/3
e x
(1
3
+ x) − e
(1 + x)2
√
3
x
We used the quotient rule and the formula (ef (x) )′ = ef (x) f ′ (x).
2. (10 points) Find the equation of the tangent line to the graph of f (x) = x2 at the point (3, 9).
.
2
Solution. We have f ′ (x) = 2x and f ′ (3) = 6. Thus the tangent line has slope 6 and its equation
is y = 6x + b. To find b, we plug in x = 3—we must get y = 9. This means we get an equation
9 = 6 · 3 + b, whence b = −9. Answer: y = 6x − 9.
3. (20 points) The graph of a function y = f (x) is given below
y
1
1
x
(a) Estimate f ′ (−3), f ′ (−2), f ′ (1).
Solution. f ′ (−3) ≈ 1, f ′ (−2) ≈ 0, f ′ (1) ≈ −2.
(b) Sketch the graph of f ′ (x) on the same picture.
(c) List the critical points of f (x).
Solution. The critical points are the points at which the tangent line is horizontal. So they are
approximately x = −2 and x = 2.5.
(d) List the intervals on which f (x) is increasing.
Solution. f (x) is increasing for −4 < x < −2 and for 2.5 < x < 4.
4. (10 points) The weight, W , in lbs, of a child is a function of its age, a, in years, so W = f (a).
(a) What are the units for f ′ (a)? The units for f ′ (a) are lbs/year.
(b) Interpret the statements f (8) = 45 and f ′ (8) = 4. This means that an 8-year old child weighs 45
lbs and is growing at a rate of 4 lbs/year.
(c) Estimate f (7) and f (10).
Solution. We have
f (7) ≈ f (8) − f ′ (8) = 45 − 4 = 41.
f (10) ≈ f (8) + 2f ′ (8) = 45 + 8 = 53.
5. (10 points) (a) The cost C (in dollars), to produce g gallons of a chemical is given by a cost
function C(g). If C(10) = 100 and the marginal cost is M C(10) = 5, estimate C(11) and C(8).
Solution.
C(11) ≈ C(10) + M C(10) = 100 + 5 = 105.
C(8) ≈ C(10) − 2M C(10) = 100 − 2 · 5 = 90.
(b) If the production level is g = 10 and the marginal revenue is M R(10) = 6 should the company
increase the production level? Why?
3
Solution. If the company increases the production level, say, by 1 gallon, the cost will increase approximately by M C(10) = 5 and the revenue will increase approximately by M R(10) = 6. Since the
increase in revenue is greater than the increase in cost, the company should increase the production
level.
6. (5 points for each part) Let f (x) = x3 − 12x + 5.
(a) Find f ′ (x).
f ′ (x) = (x3 )′ − 12(x)′ = 3x2 − 12.
(b) Find f ′′ (x).
f ′′ (x) = (3x2 − 12)′ = 3 · 2x = 6x.
(c) Find the critical points of f (x).
Solution. We solve the equation 3x2 − 12 = 0 or 3x2 = 12. Dividing by 3, we get an equation
x2 = 4. Whence
√
x = ± 4 = ±2.
There are two critical points: x = −2 and x = 2.
(d) Use the second derivative test to classify the critical points (that is, for each critical point decide
whether it is a maximum, a minimum, or neither).
Solution. We have f ′′ (−2) = 6 · (−2) = −12 < 0. Thus f (x) attains a local maximum at x = −2.
On the other hand, f ′′ (2) = 12 > 0, so f (x) attains a local minimum at x = 2.