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Math 112 (Calculus I)
Final Exam KEY
Short Answer Fill in the blank with the appropriate answer.
- (10 points)
a) d dx ln(tan x) = sec^2 x tan x
b) Use the linearization of f (x) = x^1 /^3 at a = 8 to approximate 9^1 /^3.
c) If f ′(x) = x^3 and f (0) = 5 then f (x) = x^4 4
d) If f (x) = e^2 x^ then f ′′(x) = 4e^2 x.
e) The Mean Value Theorem says that if f is a continuous function on [a, b]
which is also differentiable on (a, b) then there is a c ∈ (a, b)
with f ′(c) = f (b) − f (a) b − a
f) Circle the correct answer in both cases: If f ′^ is positive and increasing, then f is ( increasing / decreasing) and ( concave up / concave down).
g) (^) xlim→∞^ x (^3) + 2 x^3 +3x+2 =
Multiple Choice. In the grid below fill in the square corresponding to each correct answer.
2 A B C D E F G H I J
3 A B C D E F G H I J
4 A B C D E F G H I J
5 A B C D E F G H I J
6 A B C D E F G H I J
7 A B C D E F G H I J
8 A B C D E F G H I J
9 A B C D E F G H I J
10 A B C D E F G H I J
- If sin(xy) = x − 1 y , what is y′^ at (1, π)?
a) π b) π^2 − 1 c) π −
π
d) −π^2 − 1 π e) − 1 −
π f) −
π
- Which of the following has a removable discontinuity at x = 1?
a) x^2 − 1 x + 2 b) ln |x^2 − 1 | c) x^2 + 2 x − 1
d) x^2 − 3 x + 2 x^2 − 1 e) tan πx 2 f) sin
1 − x
- If the position of a particle is given by t^2 − t t − 1
, at what value(s) of t is the velocity 0?
a) 1 b) 1 ±
2 c) 2 ±
d) 1 − 2
2 e) 0 f) The velocity is never 0.
- Use one iteration of Newton’s method, beginning with x 1 = 1/2 to approximate the positive root of the equation x^2 + 2x − 1 = 0. (Note that the root is
a)
b)
c) 0
d)
e)
f) x 2 is undefined.
Free Response. For problems 11 - 19, write your answers in the space provided. Use the back of the page if needed, indicating that fact. Neatly show all work.
- (7 points) State the definition of the derivative of f (t), and use the definition to find the derivative of f (t) = 2t^2 + 1. Solution: f ′(x) = lim h→ 0 f (x + h) − f (h) h
The other definition is ok as well.
f ′(t) = lim h→ 0 2(t + h)^2 + 1 − (2t^2 + 1) h
= lim h→ 0 2 t^2 + 4th + 2h^2 − 2 t^2 h = lim h→ 0 (4t + h) = 4t.
- (7 points) Find the equation of the tangent line to y = x
x + 1 at the point (3, 6). Solution: y′^ =
x + 1 + x 2
x + 1
y′(3) = 2 +
y − 6 =
(x − 3)
y =
x −
- (7 points) Find the integral
∫ (^3) 1
x 2 x^2 + 1 dx.
Solution: Let u = 2x^2 + 1. Then du = 4x dx. The integral becomes ∫ (^19) 3
4 u du =
ln u|^13 9 = ln(19) − ln(3) 4
- (7 points) d dx xcos^ x^ = Solution: If y = xcos^ x, then ln y = cos x ln x. Differentiating, we have 1 y y′^ = − sin x ln x + cos x x
Thus, y′^ = (
cos x x − sin x ln x)xcos^ x.
- (7 points) Find the absolute minimum and maximum of f (x) = x^2 e−x (^2) / 2 on [0, 4]. Solution: f ′(x) = 2xe−x^2 /^2 − x^3 e−x^2 /^2 = x(2 − x^2 )e−x^2 /^2. Notice that there are two critical points of interest in this interval, 0 and
- Also, it is clear that on the interval (0,
- f ′^ is positive, so f is increasing. On the interval, (
2 , 4) f ′^ is negative, so f is decreasing. Hence, the maximum value of the function occurs at
2 and is 2 e
The minimum must occur at the end points. The smallest value occurs at 0, so the minimum is 0.
- (7 points) Find lim x→ 0 (1 − x)ex^ − 1 x^2
Solution:
x^ lim→ 0
(1 − x)ex^ − 1 x^2 = lim x→ 0 −ex^ + (1 − x)ex 2 x = lim x→ 0 −ex^ − ex^ + (1 − x)ex 2
- (7 points) An open-at-the-top vertical tank has horizontal cross-section of an equilateral tri- angle, and volume of 4000 cubic feet. Find the dimensions that minimize the surface area. (Hint: the height of an equilateral triangle is
times the length of a side.) Solution: The area of the equilateral base is
1 2
s · s =
s^2.
Thus, the volume of the tank is (^) √ 3 4 s^2 h = 4000. Solving for h, we have h =
3 s^2
The surface area of the tank is given by
S =
s^2 + 3sh =
s^2 + 3s ·
3 s^2
s^2 +
s
Thus, S′^ =
s −
s^2
s^3 − 32000 2 s^2
Setting S′^ = 0 we have s^3 = 32000 or s = 20 3
- Hence, h =
√^16000
3400 · 42 /^3
√^40
342 /^3
