Q. Determine the critical points of the function f(x) = x^3 - 6x^2 + 9x.
-
A.
(0, 0)
-
B.
(1, 4)
-
C.
(2, 0)
-
D.
(3, 0)
Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 1)(x - 3) = 0, so critical points are x = 1 and x = 3.
Correct Answer: D — (3, 0)
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Q. Determine the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.
-
A.
y = 3x - 2
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B.
y = 2x + 1
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C.
y = 2x + 3
-
D.
y = x + 3
Solution
f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 4). The tangent line is y - 4 = 4(x - 1) => y = 4x - 4 + 4 => y = 4x - 2.
Correct Answer: A — y = 3x - 2
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Q. Determine the intervals where the function f(x) = x^3 - 3x is increasing.
-
A.
(-∞, -1)
-
B.
(-1, 1)
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C.
(1, ∞)
-
D.
(-∞, 1)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).
Correct Answer: C — (1, ∞)
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Q. Determine the intervals where the function f(x) = x^4 - 4x^3 has increasing behavior.
-
A.
(-∞, 0) U (2, ∞)
-
B.
(0, 2)
-
C.
(0, ∞)
-
D.
(2, ∞)
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). The function is increasing where f'(x) > 0, which is in the intervals (-∞, 0) and (3, ∞).
Correct Answer: A — (-∞, 0) U (2, ∞)
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Q. Determine the local maxima and minima of f(x) = x^3 - 3x.
-
A.
Maxima at (1, -2)
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B.
Minima at (0, 0)
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C.
Maxima at (0, 0)
-
D.
Minima at (1, -2)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f''(1) = 6 > 0 (min), f''(-1) = 6 > 0 (min). Local maxima at (0, 0).
Correct Answer: A — Maxima at (1, -2)
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Q. Determine the local maxima and minima of the function f(x) = x^3 - 6x^2 + 9x.
-
A.
(0, 0)
-
B.
(2, 0)
-
C.
(3, 0)
-
D.
(1, 0)
Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f''(1) > 0 (min), f''(3) < 0 (max).
Correct Answer: C — (3, 0)
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Q. Determine the local maxima and minima of the function f(x) = x^4 - 4x^3 + 4x.
-
A.
Maxima at (0, 0)
-
B.
Minima at (2, 0)
-
C.
Maxima at (2, 0)
-
D.
Minima at (0, 0)
Solution
f'(x) = 4x^3 - 12x^2 + 4. Setting f'(x) = 0 gives x = 0 and x = 2. f''(0) = 4 > 0 (min), f''(2) = -8 < 0 (max).
Correct Answer: B — Minima at (2, 0)
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Q. Determine the maximum value of f(x) = -x^2 + 4x + 1.
Solution
The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5, which is the maximum value.
Correct Answer: B — 5
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Q. Determine the minimum value of the function f(x) = x^2 - 4x + 5.
Solution
The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1. Thus, the minimum value is 1.
Correct Answer: A — 1
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Q. Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local minimum.
-
A.
(1, 2)
-
B.
(2, 1)
-
C.
(0, 4)
-
D.
(3, 4)
Solution
Find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, 1) is a local minimum.
Correct Answer: A — (1, 2)
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Q. Determine the point of inflection for the function f(x) = x^4 - 4x^3 + 6.
-
A.
(1, 3)
-
B.
(2, 2)
-
C.
(3, 1)
-
D.
(0, 6)
Solution
f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x = 0 and x = 2. The point of inflection is at (1, 3).
Correct Answer: A — (1, 3)
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Q. Determine the point of inflection for the function f(x) = x^4 - 4x^3 + 6x^2.
-
A.
(1, 3)
-
B.
(2, 2)
-
C.
(3, 1)
-
D.
(0, 0)
Solution
Find f''(x) = 12x^2 - 24x + 12. Setting f''(x) = 0 gives x = 1 and x = 2. Testing intervals shows a change in concavity at x = 1.
Correct Answer: A — (1, 3)
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Q. Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the slope of the tangent is 0.
-
A.
(1, 0)
-
B.
(0, 2)
-
C.
(2, 0)
-
D.
(3, 2)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x^2 = 1, so x = 1 or x = -1. f(1) = 0, f(-1) = 4. The point is (1, 0).
Correct Answer: A — (1, 0)
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Q. Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the tangent is horizontal.
-
A.
(0, 2)
-
B.
(1, 0)
-
C.
(2, 0)
-
D.
(3, 2)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. The point is (1, 0).
Correct Answer: B — (1, 0)
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Q. Find the coordinates of the point where the function f(x) = 3x^2 - 12x + 9 has a local maximum.
-
A.
(2, 3)
-
B.
(3, 0)
-
C.
(1, 1)
-
D.
(0, 9)
Solution
f'(x) = 6x - 12. Setting f'(x) = 0 gives x = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.
Correct Answer: A — (2, 3)
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Q. Find the critical points of the function f(x) = 3x^4 - 8x^3 + 6.
-
A.
(0, 6)
-
B.
(2, -2)
-
C.
(1, 1)
-
D.
(3, 0)
Solution
f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x^2(12x - 24) = 0, so x = 0 or x = 2. f(2) = 3(2^4) - 8(2^3) + 6 = -2.
Correct Answer: B — (2, -2)
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Q. Find the critical points of the function f(x) = x^3 - 6x^2 + 9x.
-
A.
(0, 0)
-
B.
(3, 0)
-
C.
(2, 0)
-
D.
(1, 0)
Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1 and x = 3. Critical points are (1, f(1)) and (3, f(3)).
Correct Answer: B — (3, 0)
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Q. Find the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.
-
A.
y = 3x - 2
-
B.
y = 2x + 1
-
C.
y = 2x + 2
-
D.
y = x + 3
Solution
f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 3). The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.
Correct Answer: A — y = 3x - 2
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Q. Find the intervals where the function f(x) = x^4 - 4x^3 has increasing behavior.
-
A.
(-∞, 0)
-
B.
(0, 2)
-
C.
(2, ∞)
-
D.
(0, 4)
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). Critical points are x = 0 and x = 3. Test intervals: f' is positive in (0, 3) and (3, ∞).
Correct Answer: B — (0, 2)
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Q. Find the maximum value of f(x) = -x^2 + 4x + 1.
Solution
The maximum occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5.
Correct Answer: A — 5
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Q. Find the maximum value of the function f(x) = -2x^2 + 8x - 3.
Solution
The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.
Correct Answer: B — 8
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Q. Find the maximum value of the function f(x) = -2x^2 + 8x - 5.
Solution
The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = -2(2^2) + 8(2) - 5 = 9.
Correct Answer: C — 9
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Q. Find the maximum value of the function f(x) = -x^2 + 4x + 1.
Solution
The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 9, which is the maximum value.
Correct Answer: A — 5
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Q. Find the maximum value of the function f(x) = -x^2 + 6x - 8.
Solution
The vertex occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.
Correct Answer: C — 8
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Q. Find the minimum value of the function f(x) = 3x^2 - 12x + 7.
Solution
The vertex occurs at x = -b/(2a) = 12/6 = 2. f(2) = 3(2^2) - 12(2) + 7 = 1.
Correct Answer: B — 1
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Q. Find the minimum value of the function f(x) = x^2 - 4x + 5.
Solution
The vertex of the parabola occurs at x = 2. f(2) = 2^2 - 4(2) + 5 = 1, which is the minimum value.
Correct Answer: A — 1
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Q. Find the minimum value of the function f(x) = x^4 - 8x^2 + 16.
Solution
f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0, which is the minimum value.
Correct Answer: A — 0
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Q. Find the point of inflection for the function f(x) = x^3 - 6x^2 + 9x.
-
A.
(1, 4)
-
B.
(2, 3)
-
C.
(3, 0)
-
D.
(0, 0)
Solution
f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. The point of inflection is (2, f(2)) = (2, 3).
Correct Answer: C — (3, 0)
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Q. Find the point of inflection for the function f(x) = x^4 - 4x^3 + 6.
-
A.
(1, 3)
-
B.
(2, 2)
-
C.
(3, 1)
-
D.
(0, 6)
Solution
f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. The point of inflection is at (2, f(2)) = (2, 2).
Correct Answer: A — (1, 3)
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Q. Find the slope of the tangent line to the curve y = sin(x) at x = π/4.
-
A.
1
-
B.
√2/2
-
C.
√3/2
-
D.
0
Solution
The derivative f'(x) = cos(x). At x = π/4, f'(π/4) = cos(π/4) = √2/2.
Correct Answer: B — √2/2
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