Multiple Choice
Question 1
A function is concave up on an interval when:
A
f''(x) > 0
B
f'(x) > 0
C
f(x) > 0
D
f''(x) < 0
Solution: A positive second derivative means the function is concave up (its slope is increasing).
Question 2
An inflection point of f occurs where:
A
concavity changes (f'' changes sign)
B
f' = 0
C
f = 0
D
f has a maximum
Solution: An inflection point is where the graph changes concavity, which requires f'' to change sign.
Question 3
If f'(x) < 0 on an interval, then on that interval f is:
A
decreasing
B
increasing
C
constant
D
concave up
Solution: A negative first derivative means the function is decreasing.
Question 4
At a local extremum of a differentiable function, the first derivative is:
A
zero
B
positive
C
negative
D
undefined
Solution: For a differentiable function, local extrema occur where f'(x) = 0.
Question 5
The conclusion of the Mean Value Theorem is that there exists c in (a, b) such that:
A
f'(c) = (f(b) - f(a))/(b - a)
B
f'(c) = 0
C
f(c) = 0
D
f''(c) = 0
Solution: The MVT concludes that some interior point c has instantaneous rate f'(c) equal to the average rate of change over [a, b].
Question 6
An absolute extremum on a closed interval is:
A
the single largest (or smallest) function value on the whole interval
B
any local extremum
C
always at the midpoint
D
where f'' = 0
Solution: An absolute (global) extremum is the largest or smallest value the function attains over the entire interval.
Question 7
By the First Derivative Test, f has a local maximum at c if f' changes from:
A
positive to negative at c
B
negative to positive at c
C
positive to positive
D
zero to zero
Solution: A local maximum occurs where the function stops increasing and starts decreasing, i.e., f' changes from positive to negative.
Question 8
For f(x) = x^3 - 3x, at which x is there a local maximum?
A
x = -1
B
x = 1
C
x = 0
D
x = 3
Solution: f'(x) = 3(x-1)(x+1). At x = -1, f' changes from positive (x < -1) to negative (-1 < x < 1), a local maximum.
Question 9
For f(x) = x^3 - 3x, at which x is there a local minimum?
A
x = 1
B
x = -1
C
x = 0
D
x = 2
Solution: At x = 1, f' changes from negative to positive, indicating a local minimum.
Question 10
How many critical numbers does f(x) = x^4 - 4x^2 have?
A
3
B
1
C
2
D
4
Solution: f'(x) = 4x^3 - 8x = 4x(x^2 - 2) = 0 gives x = 0, x = sqrt(2), x = -sqrt(2): three critical numbers.
Question 11
If f'(x) = x^2 (does not change sign at x = 0), what happens at x = 0?
A
There is no local extremum (f' does not change sign)
B
Local maximum
C
Local minimum
D
Inflection point guaranteed
Solution: Although x = 0 is a critical number, f'(x) = x^2 >= 0 on both sides, so f' does not change sign and there is no local extremum.
Question 12
Find the absolute maximum of f(x) = x^2 on the interval [-1, 3].
A
9
B
1
C
0
D
3
Solution: Candidates: critical point x = 0 (f = 0) and endpoints x = -1 (f = 1), x = 3 (f = 9). The absolute maximum is 9.
Question 13
Find the absolute minimum of f(x) = x^2 - 4x + 1 on [0, 3].
A
-3
B
1
C
-2
D
0
Solution: f'(x) = 2x - 4 = 0 at x = 2. f(2) = 4 - 8 + 1 = -3, f(0) = 1, f(3) = 9 - 12 + 1 = -2. The absolute minimum is -3 at x = 2.
Question 14
The 'candidates test' for absolute extrema on [a, b] evaluates f at:
A
all critical points in [a, b] and both endpoints
B
only the midpoint
C
only inflection points
D
only where f'' = 0
Solution: To find absolute extrema on a closed interval, compare f-values at every critical point and at both endpoints.
Question 15
If f'(x) = 0 for all x in an interval, then on that interval f is:
A
constant
B
increasing
C
decreasing
D
concave up
Solution: A consequence of the MVT: if the derivative is zero everywhere on an interval, the function is constant there.
Question 16
If two functions f and g satisfy f'(x) = g'(x) for all x, then:
A
f(x) = g(x) + C for some constant C
B
f(x) = g(x)
C
f(x) = C g(x)
D
f and g are unrelated
Solution: If two functions have equal derivatives on an interval, they differ by a constant: f(x) - g(x) = C.
Question 17
For f(x) = x^3 - 3x, use the Second Derivative Test at x = 1.
A
Local minimum, since f''(1) = 6 > 0
B
Local maximum
C
Inflection point
D
No conclusion
Solution: f''(x) = 6x, so f''(1) = 6 > 0. The function is concave up at x = 1, giving a local minimum.
Question 18
For f(x) = x^3 - 3x, apply the Second Derivative Test at x = -1.
A
Local maximum, since f''(-1) = -6 < 0
B
Local minimum
C
Inflection point
D
No conclusion
Solution: f''(-1) = 6(-1) = -6 < 0. The function is concave down, giving a local maximum.
Question 19
For f(x) = x^2 + 4x, the Second Derivative Test at the critical point x = -2 shows:
A
a local minimum (f'' = 2 > 0)
B
a local maximum
C
an inflection point
D
no extremum
Solution: f'(x) = 2x + 4 = 0 at x = -2. f''(x) = 2 > 0, so x = -2 is a local minimum.
Question 20
For f(x) = x^3, on what interval is f concave down?
A
x < 0
B
x > 0
C
all x
D
never
Solution: f''(x) = 6x < 0 when x < 0, so f is concave down for x < 0.