Math 220, Sections 04**, T Pilachowski, Spring 2007
Spring 2007 MATH 220 – TEST 1(A) (0.3 – 2.7) [Pilachowski] ANSWER KEY
1. a. (10 points) Given
()
4
2
3+−= x
xxf , find
(
)
2
−
′
f.
() ( ) ( ) ()
2
25
2
1
12
2
2
2322342 2
2
2213 =+=
−
+−=−
′
⇒+=
′
⇒+−= −− fxxfxxxf
b. (4 extra points) Determine the equations of all asymptotes for
()
2
3
5−−= x
xxp .
vertical asymptote is x = 0; As
(
)
205,
−
−
→∞→ xxpx : slant asymptote is y = 5x – 2
c. (10 points) Given
() ( )
3
52 −= xxg , use calculus to determine whether the curve is concave up,
concave down or has a point of inflection at x = 0.
() () () () ()
⇒<
′′
⇒−=∗−=
′′
⇒−=∗−=
′005224252125262523 22 gxxgxxg concave down
d. (18 points) Xanthu wants to fence in a rectangular garden and has $1200 to spend. One side needs a
stronger fence—materials will cost $4 per foot for this side. The other three sides can be fenced with
less expensive materials at $2 per foot. Use calculus to find the dimensions which will maximize area,
and indicate which value applies to the stronger side.
constraint: 4x + 2x + 2(2y)=1200 xy 2
3
300 −=⇒ ; objective: xAxxxyA 3300
2
3
300 2−=
′
⇒−==
300 – 3x = 0 ⇒ x = 100 and y = 150; strong side is 100 ft and other dimension is 150 ft.
2. a. (12 points) Given
()
23
3
1
4
1234 +−−= xxxxh , use calculus to find values for x for which the
function is increasing, and values for x for which the function is decreasing.
() ( )( )
0326
23 =−+=−−=
′xxxxxxxh for x = –2, 0, 3; for x < –2, h
′
< 0; for –2 < x < 0, h
′
> 0;
for 0 < x < 3, h′ < 0; for 3 < x, h′ > 0;
h is increasing for –2 < x < 0 and x > 3; h is decreasing for x < –2 and 0 < x < 3;
b. (12 points) Given
()
52122 234 −+−+= xxxxxm , use calculus to find values for x for which the
graph of m has a point of inflection.
() () ( )( )
01212241212,22464 223 =−+=−+=
′′
+−+=
′xxxxxmxxxxm for x = –2, 1
x < –2 –2 < x < 1 1 < x
m′′ positive negative positive
3. a. (10 points) Given a company’s Revenue function
()
x
x
xR −
+
−= 2
400
500 , find the equation to
express Marginal Revenue.
() () ()
124002400500 21 −+=
′
⇒−+−= −− xRxxxR
b. (10 points) Given
()
13 += xxp , find average rate of change for p for 1 ≤ x ≤ 5. You must show
your work and state a numeric answer in simplest form.
() ()
41155,2131 =+==+= pp , average rate of change =
(
)
(
)
2
1
4
24
15
15 =
−
=
−
−
pp
The graph has points of inflection at both x = –2 and x = 1.
