104/184 Quiz #4
November 21
Grade:
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Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Below is a graph of f 00 (x).
(a) Where is f (x) concave up?
Answer: (−1, 0)
S
(3, ∞)
(b) Given that f 0 (1) = 0, is there a local maximum or minumum when x = 1?
Answer: f 00 (1) < 0 ⇒ f (x) is concave
down ⇒ local max
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
√
(a) The function f (x) = x · (x − 1) has a critical point at x = 1/3. Determine whether it is
local maximum or minimum. Justify your answer!
Answer: local minimum
√
√
√ . Since f 0 (x) changes from
Solution: f 0 (x) = 2√1 x · (x − 1) + x = x−1+2x
= 3x−1
2 x
2 x
negative to positive at x = 1/3 there is local minimum there.
(b) Determine where
f (x) =
x2 − 9
x(x − 3)
has a vertical asymptote.
Answer: VA at x = 0.
2
x −9
Solution: f (x) = x(x−3)
= (x−3)(x+3)
=
x(x−3)
asymptote at x = 0 (but not at x = 3).
x+3
x
from which we see there is a vertical
Long answer question — you must show your work
3. 4 marks Sketch the graph of the function
f (x) = (x − 2)2 (x + 2)
given that:
f 0 (x) = (3x + 2)(x − 2) and f 00 (x) = 6x − 4.
Solution: Note that f 0 (x) is a parabola opening upward with roots x = −2/3 and x − 2.
Therefore, f 0 (x) < 0 when −2/3 < x < 2 and f 0 (x) > 0 when x < −2/3 or x > 2. Consequently, there must be a local max at x = −2/3 and a local min at x = 2.
x-intercepts: (2, 0) and (−2, 0). y-intercept: (0, 8)
It’s also pretty easy to see that f 00 (x) < 0 when x < 4/6 = 2/3 and f 00 (x) > 0 when
x > 4/6 = 2/3. Hence, there is an inflection point at x = 2/3.
The graph should look something like this: