Multiple Choice
Question 1
For f(x) = x^4, describe the concavity.
A
Concave up for all x (except possibly x = 0)
B
Concave down everywhere
C
Concave up for x > 0 only
D
Never concave up
Solution: f''(x) = 12x^2 >= 0, and is positive for all x not equal to 0, so f is concave up.
Question 2
Concave up means the graph:
A
opens upward and lies above its tangent lines
B
opens downward
C
is a straight line
D
is decreasing
Solution: A concave-up curve bends upward (like a cup) and lies above each of its tangent lines.
Question 3
If f'(x) > 0 and f''(x) < 0 on an interval, the graph of f is:
A
increasing and concave down
B
decreasing and concave up
C
increasing and concave up
D
decreasing and concave down
Solution: f' > 0 means increasing; f'' < 0 means concave down. So the graph rises while bending downward.
Question 4
If the graph of f' is increasing on an interval, then f is:
A
concave up
B
concave down
C
decreasing
D
constant
Solution: f' increasing means f'' > 0, which means f is concave up.
Question 5
If f'(c) = 0 and f' changes from negative to positive at c, then the graph of f has:
A
a local minimum at c
B
a local maximum at c
C
an inflection point at c
D
a vertical asymptote
Solution: f' going from negative to positive means f goes from decreasing to increasing: a local minimum.
Question 6
Find the inflection point of f(x) = x^3.
A
x = 0
B
x = 1
C
x = -1
D
no inflection point
Solution: f''(x) = 6x = 0 at x = 0, and f'' changes sign there (negative to positive), so x = 0 is an inflection point.
Question 7
For f(x) = x^3 - 6x^2 + 5, at what x is the inflection point?
A
x = 2
B
x = 0
C
x = 6
D
x = 5
Solution: f''(x) = 6x - 12 = 0 at x = 2, and concavity changes there, so x = 2 is an inflection point.
Question 8
The number of inflection points of f(x) = x^4 - 6x^2 is:
A
2
B
1
C
0
D
3
Solution: f''(x) = 12x^2 - 12 = 12(x^2 - 1) = 0 at x = 1 and x = -1, with sign changes at both, giving 2 inflection points.
Question 9
The Intermediate Value Theorem guarantees a solution to f(x) = k on [a, b] if f is continuous and:
A
k is between f(a) and f(b)
B
k > f(b)
C
f is increasing
D
f'(x) = 0 somewhere
Solution: The IVT guarantees that a continuous function takes every value k between f(a) and f(b) at some point in [a, b].
Question 10
Which theorem guarantees a continuous function on [a, b] attains its maximum value?
A
Extreme Value Theorem
B
Mean Value Theorem
C
Intermediate Value Theorem
D
Rolle's Theorem
Solution: The Extreme Value Theorem guarantees that a function continuous on a closed interval attains an absolute maximum (and minimum).
Question 11
A rectangle has a fixed perimeter of 40. What dimensions maximize its area?
A
10 by 10 (a square)
B
5 by 15
C
8 by 12
D
1 by 19
Solution: With perimeter 40, l + w = 20. Area A = w(20 - w) is maximized at w = 10, giving a 10 by 10 square with area 100.
Question 12
What is the minimum value of the sum S = x + 16/x for x > 0?
A
8
B
16
C
4
D
12
Solution: S'(x) = 1 - 16/x^2 = 0 at x = 4. S(4) = 4 + 4 = 8, a minimum since S'' > 0.
Question 13
The first step in solving an optimization problem is usually to:
A
write the quantity to optimize as a function of one variable
B
take the second derivative
C
guess the answer
D
integrate
Solution: You express the objective (e.g., area or cost) as a function of a single variable using the constraint, then find critical points.
Question 14
The difference between an absolute maximum and a relative (local) maximum is:
A
an absolute maximum is the highest value on the whole domain; a relative maximum is highest only nearby
B
there is no difference
C
a relative maximum is always higher
D
an absolute maximum occurs only at endpoints
Solution: An absolute (global) maximum is the largest value over the entire domain, while a relative (local) maximum is the largest value only in a neighborhood.
Question 15
A relative extremum of a differentiable function can only occur where:
A
f'(x) = 0
B
f(x) = 0
C
f''(x) = 0
D
x = 0
Solution: For a differentiable function, relative extrema occur only at critical points where f'(x) = 0 (Fermat's Theorem).
Question 16
If f'(x) = -2x + 6, on what interval is f increasing?
A
x < 3
B
x > 3
C
all x
D
x < 0
Solution: f'(x) = -2x + 6 > 0 when x < 3, so f is increasing for x < 3.
Question 17
For f(x) = 3x^4 - 4x^3, the critical numbers are:
A
x = 0 and x = 1
B
x = 1 only
C
x = 0 only
D
x = -1 and x = 1
Solution: f'(x) = 12x^3 - 12x^2 = 12x^2(x - 1) = 0 at x = 0 and x = 1.
Question 18
For f(x) = -x^2 + 6x, describe the concavity everywhere.
A
Concave down for all x
B
Concave up for all x
C
Concave up for x > 3
D
Changes concavity at x = 3
Solution: f''(x) = -2 < 0 for all x, so the parabola is concave down everywhere.
Question 19
Find the absolute maximum of f(x) = 2x^3 - 3x^2 - 12x on [-2, 3].
A
7
B
20
C
-20
D
0
Solution: f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1) = 0 at x = -1, 2. f(-2) = -16 -12 +24 = -4, f(-1) = -2 -3 +12 = 7, f(2) = 16 -12 -24 = -20, f(3) = 54 -27 -36 = -9. Absolute maximum is 7 at x = -1.
Question 20
A function continuous on [2, 8] is guaranteed by the EVT to have:
A
both an absolute maximum and an absolute minimum on [2, 8]
B
no extrema
C
an inflection point
D
a vertical asymptote
Solution: The EVT guarantees that a function continuous on a closed, bounded interval attains both an absolute maximum and an absolute minimum.