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MATH 145
WINTER 2008
SAMPLE FINAL EXAM — HINTS AND ANSWERS
1. ANSWERS:
(a)
dy dx = cos(cos(ln^ x))^ ·^ (−^ sin(ln^ x))^ ·^
x
(b) dy dx
=^1
sin x xe^3 x−^5
)− 1 / 2 [
xe^3 x−^5 · cos x − (sin x)(xe^3 x−^5 (3) + e^3 x−^5 ) (xe^3 x−^5 )^2
]
(c)
dy dx =
(x^2 + 5) · 3(tan^2 x)(sec^2 x) − tan^3 x(2x) (x^2 + 5)^2
(d) dy dx
= cos(x
x + 1)
[
x · 1 2
x + 1
x + 1 · (1)
]
(e)
dy dx =
(x^3 + 2x) · etan^ x^ · (sec^2 x) − etan^ x(3x^2 + 2) (x^3 + 2x)^2
(f) dy dx
sin(x^2 )
[
− sin
x^2
)] (
x^3
x^2
cos(x^2 )
- (a) ANSWER: f ′(x) = − 2 xe−x^2 and f ′′(x) = 2e−x^2 (2x^2 − 1) (simplified) (b) HINT: f ′(x) = 0 only at x = 0. Set up a sign chart to show that f (x) has a relative maximum at x = 0. ANSWER: The only critical point is at x = 0 and this gives a relative maximum.
(c) HINT: f ′′(x) = 0 at x = ±
. So, f (x) has possible inflection points at x = ±
Set up a sign chart to show that f ′′(x) changes sign at each of these values of x.
ANSWER: f (x) has inflection points at x = −
2 and^ x^ =
(d) ANSWER: The y-intercept is f (0) = 1. Since f (x) is never 0, there is no x-intercept. (e) HINT: To get the horizontal asymptote, evaluate (^) xlim→∞ e−x
2 . ANSWER: There are no vertical asymptotes. The line y = 0 (the x-axis) is the horizontal asymptote. (f) You should get a nice bell-shaped curve.
(0, 1) = relative maximum and y-intercept
1 2 , e−^1 /^2
= i.p.
1 2 , e−^1 /^2
= i.p.
- (a) HINT: Use substitution with u = ln 4x. ANSWER: 14 (ln 4x)^2 + C (b) HINT: Use substitution with w = 2x + 4. Then use integration by parts to do the resulting integral. ANSWER: 12 (2x + 4) ln(2x + 4) − 12 (2x + 4) + C (Your answer may differ from this one by a constant.)
(c) ANSWER: − 252 (d) ANSWER: (^283) (e) HINT: Use substitution with u = 3x + 5. Then do i.b.p. with w = ln u and dv = du. ANSWER: 13 (3x + 5) ln(3x + 5) − 3 x 3 +5 + C (f) HINT: Do substitution with u = x^2 − 3 x + 1. ANSWER: − 2
x^2 − 3 x + 1 + C (g) HINT: Do substitution with u = x^6. ANSWER: 23 tan(x^6 ) + C (h) HINT: Do substitution with u = 1 + cos x. ANSWER: ln 2 (i) HINT: Do substitution with u = sin x. Then do partial fractions on the resulting integral. ANSWER: ln | sin x| − ln | 1 − sin x| − (^) sin^1 x + C
- (a) ANSWER: diverges
(b) ANSWER: converges to 1
(c) HINT: Split the integral into two integrals:
−∞
e−|x|^ dx and
0
e−|x|, dx. Then use the fact that |x| = −x if x is negative and |x| = x if x is positive. ANSWER: converges to 2
- ANSWER: $
- HINT: Let x be the length of the side of the square bottom/top and let y be the height of the box. Then the volume of the box is x^2 y, which must be equal to 320. The cost of the box is 0. 1 x^2 + 0. 08 x^2 + 4(0. 2 xy). Solve the equation x^2 y = 320 for y and substitute into the cost formula to get a cost function of a single variable: C(x) = 0. 18 x^2 + (^256) x. Find the value of x that minimizes cost. ANSWER: x = 8.93 and y = 4. 02
- HINT: Convert all of the values of X to values on the standard normal curve by using the change of variable Z = X−σ μ.
(a) HIINT: P (X ≥ 6) = P (Z ≥ − 2 .33) = P (Z ≤ 2 .33) ANSWER: 99.01% (b) HINT: P (X ≤ 8) = P (Z ≤ − 1 .67) ANSWER: 0. (c) HINT: P (13 ≤ X ≤ 15) = P (0 ≤ Z ≤ 0 .67) ANSWER: 24.86%
- (a) HINT: Evaluate the integral
0
4 xe−^4 x^ dx.
(b) HINT: P (0 ≤ X ≤ 5) =
0 4 e−^4 x^ dx ANSWER: 1 − e−^20 = 0. 9999999999999
- HINT: You need
∫ (^) b
0
x^2 e−^
(^14) x 3 dx = 1.
ANSWER: b = 3
−4 ln( 14 )
