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Math 112 (Calculus I)
Final Exam Form A KEY
Multiple Choice. Fill in the answer to each problem on your scantron. Make sure your name, section and instructor is on your scantron.
- Simplify sin(tan−^1
a) 1 b)
c)
d)
e)
f) 3
g) None of the above. Solution: d)
- Find lim x→ 1 x^2 + x + 2 2 x^2 + x − 3 a) Does not exist b) -1 c) 2
d)
e) 4 f)
g) None of the above. Solution: a)
- If f (x) = Ax^2 + Bx + C, find lim h→ 0
f (x + h) − f (x) h
a) Ax^2 + Bx b) 2 Ax + B c) 2 Ax^2 + Bx + C
d) 2 Ax^2 + Bx e) 0 f) Does not exist.
Solution: b)
- The derivative of h(x) = x^2 x^ is a) 2 x^2 x−^1 b) 2 x · x^2 x−^1 c) x^2 x^ ln x
d) 2 x^2 x(1 + ln x) e) 2 x^2 x
Solution: d)
- The equation of the tangent line to y = sin^2 x
at ( π 6
) is
a) y =
- 2(x − π 6 ) sin x cos x b) y =
x + 3 − π
c) y =
d) y =
x + 3 − π 12
e) None of the above.
- For the function f (x) = x^2 , for which value of c is f ′(c) equal to the average rate of change of f on [0, 3]?
a) 0 b)
c) 3
d)
e) 6 f) There is no such c.
g) None of the above.
Solution: b)
- A farmer wishes to fence in 100 square feet of area into two adjoining rectangular regions. The outside fence costs $8 per linear foot and the dividing fence inside costs $9 per linear foot. What is the minimal cost to fence the region?
a) $ 100 b) $ 200 c) $ 300
d) $ 400 e) $ 600 f) None of the above.
Solution: d)
∫ (^1) 0
x
1 + 3x^2 dx =
a)
b)
c)
d) 7 e) 10 f) None of the above.
Solution: b)
Free Response. For problems 10 - 17, write your answers in the space provided. Use the back of the page if needed, indicating that fact. Neatly show all work.
- (8 points)
(a) State the definition of the derivative. Solution: Either f ′(a) = lim x→a
f (x) − f (a) x − a or f ′(x) = lim h→ 0
f (x + h) − f (x) h is acceptable.
(b) Find f ′(2) where f (x) =
x − 1 using the definition of the derivative. Solution: Version A:
f ′(2) = lim x→ 2
1 x− 1 −^1 x − 2
= lim x→ 2 1 − (x − 1) (x − 2)(x − 1)
= lim x→ 2
x − 1
Version B:
f ′(−1) = (^) xlim→− 1
1 x− 1 −^ (
− 1 2 ) x − (−1)
= (^) xlim→− 1 2 + (x − 1) 2(x − 1)(x + 1)
= (^) xlim→− 1
2(x − 1)
- (8 points) Given the equation x^6 + 3xy + y^5 = 5, find y′^ at the point (1, 1). Solution: We differentiate implicitly: Version A:
6 x^5 + 3y + 3xy′^ + 5y^4 y′^ = 0 Substituting 1 for x and 1 for y, we have
9 + 8y′^ = 0,
or y′^ = −
Version B: 6 x^5 + 2y + 2xy′^ + 10y^4 y′^ = 0 8 + 12y′^ = 0
y′^ = −
- (8 points) The volume of a sphere is increasing at a rate of 3 cubic centimeters per second. How fast is the radius increasing when the radius is 10 centimeters? Solution:
V =
πr^3 dV dt = 4πr^2 dr dt Version A:
3 = 4π(100) dr dt dr dt
400 π cm/s
Version B:
2 = 4π(81) dr dt dr dt
162 π cm/s
- (8 points) Find all local extrema of the function
f (x) = (x^2 − 3)ex,
and classify them as local maxima or local minima. Solution:
f ′(x) = 2xex^ + (x^2 − 3)ex^ = (x^2 + 2x − 3)ex^ = (x + 3)(x − 1)ex.
Thus, x = −3 and x = 1 are critical points.
f ′′(x) = (2x + 2)ex^ + (x^2 + 2x − 3)ex^ = (x^2 + 4x − 1)ex. f ′′(−3) = (9 − 12 − 1)e−^3 < 0 f ′′(1) = (1 + 4 − 1)e^1 > 0. Thus, f has a maximum of f (−3) = 6e−^3 and a minimum of f (1) = − 2 e.
- (9 points) Sketch the curve of
f (x) = x − 1 x^2
labelling all asymptotes, intercepts, local max and mins, and inflection points. Solution: Note: This problem can also be done by rewriting the function as
f (x) = x−^1 − x−^2.
the solution will be given without the simplification in case some students do it that way. Clearly, the function has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. Also, (1, 0) is the only intercept.
f ′(x) = x^2 − 2 x(x − 1) x^4
2 − x x^3
The only critical point, therefore is x = 2.
f ′′(x) = −x^3 − (2 − x)3x^2 x^6
2 x − 6 x^4
Notice that f ′′(2) < 0, so we have a local maximum at (2,
). Also, f ′′(3) = 0. Since f ′′(−1) < 0, f is concave down on (−∞, 0). As f ′′(2) < 0, f is concave down on (0, 3). As f ′′(4) > 0, f is concave up on (3, ∞). Thus, (3,
) is an inflection point.
END OF EXAM
