Q. If the function f(x) = e^x + x^2 has a minimum at x = 0, then f(0) is:
Solution
Evaluating f(0) = e^0 + 0^2 = 1 + 0 = 1.
Correct Answer: A — 1
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Q. The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:
-
A.
x = 1, 3
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B.
x = 0, 2
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C.
x = 2, 4
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D.
x = 1, 2
Solution
Finding f'(x) = 3x^2 - 12x + 9 and solving gives critical points at x = 1 and x = 3.
Correct Answer: A — x = 1, 3
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Q. The equation of the tangent line to the curve y = x^2 at the point (2, 4) is:
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A.
y = 2x
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B.
y = 4x - 4
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C.
y = 4x - 8
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D.
y = x + 2
Solution
The slope of the tangent at x = 2 is f'(x) = 2x, so f'(2) = 4. The equation of the tangent line is y - 4 = 4(x - 2), which simplifies to y = 4x - 8.
Correct Answer: C — y = 4x - 8
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Q. The equation of the tangent to the curve y = x^2 at the point (2, 4) is:
-
A.
y = 2x - 4
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B.
y = 2x
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C.
y = x + 2
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D.
y = x^2 - 2
Solution
The derivative f'(x) = 2x. At x = 2, f'(2) = 4. The equation of the tangent line is y - 4 = 4(x - 2), which simplifies to y = 2x - 4.
Correct Answer: A — y = 2x - 4
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Q. The function f(x) = ln(x) + x has a minimum at:
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A.
x = 1
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B.
x = 0
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C.
x = e
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D.
x = 2
Solution
Finding f'(x) = 1/x + 1. Setting f'(x) = 0 gives x = 1 as the minimum point.
Correct Answer: A — x = 1
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Q. The function f(x) = x^3 - 6x^2 + 9x has how many local extrema?
Solution
Finding f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1 and x = 3. Checking the second derivative shows one local maximum and one local minimum.
Correct Answer: B — 1
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Q. The maximum value of the function f(x) = -x^2 + 4x + 1 is at x = ?
Solution
To find the maximum, we calculate f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, this is a maximum point.
Correct Answer: B — 2
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Q. The maximum value of the function f(x) = -x^2 + 4x + 1 is:
Solution
The vertex form of a parabola gives the maximum value at x = -b/(2a) = 2. Evaluating f(2) = -2^2 + 4*2 + 1 = 9.
Correct Answer: A — 5
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Q. The maximum value of the function f(x) = -x^2 + 4x + 1 occurs at:
-
A.
x = 2
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B.
x = 4
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C.
x = 1
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D.
x = 3
Solution
The vertex of the parabola given by f(x) = -x^2 + 4x + 1 occurs at x = -b/(2a) = -4/(-2) = 2, which gives the maximum value.
Correct Answer: A — x = 2
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Q. The minimum value of the function f(x) = x^4 - 8x^2 + 16 is:
Solution
Finding the derivative f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. Evaluating f(0) = 16, f(2) = 0, and f(-2) = 0, the minimum value is 0.
Correct Answer: A — 0
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Q. The slope of the tangent to the curve y = sin(x) at x = π/4 is:
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A.
1
-
B.
√2/2
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C.
√3/3
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D.
√2
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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Q. The slope of the tangent to the curve y = x^3 - 3x at x = 1 is:
Solution
The derivative f'(x) = 3x^2 - 3. At x = 1, f'(1) = 3(1)^2 - 3 = 0, so the slope is 0.
Correct Answer: B — 1
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Q. What is the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1?
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A.
y = 3x - 2
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B.
y = 2x + 1
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C.
y = 2x + 2
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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. What is the equation of the tangent line to the curve y = x^2 + 2x at the point (1, 3)?
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A.
y = 2x + 1
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B.
y = 2x + 2
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C.
y = 3x
-
D.
y = x + 2
Solution
f'(x) = 2x + 2. At x = 1, f'(1) = 4. The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.
Correct Answer: A — y = 2x + 1
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Q. What is the maximum slope of the function f(x) = -x^2 + 4x?
Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. The maximum slope occurs at x = 2, f'(2) = 2.
Correct Answer: A — 4
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Q. What is the maximum value of the function f(x) = -x^2 + 4x + 1?
Solution
The vertex occurs at x = -b/(2a) = 4/(-2) = 2. f(2) = -2^2 + 4*2 + 1 = 6.
Correct Answer: B — 6
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Q. What is the minimum value of the function f(x) = 2x^2 - 8x + 10?
Solution
The vertex form gives the minimum at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.
Correct Answer: B — 4
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Q. What is the minimum value of the function f(x) = x^2 - 4x + 5?
Solution
The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.
Correct Answer: A — 1
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Q. What is 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(x^2 - 4) = 0, so x = 0, 2, -2. f(0) = 16, f(2) = 0, f(-2) = 0. Minimum value is 0.
Correct Answer: A — 0
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Q. What is the slope of the tangent line to the curve y = x^2 + 2x at the point (1, 3)?
Solution
The derivative y' = 2x + 2. At x = 1, y' = 2(1) + 2 = 4, which is the slope of the tangent line.
Correct Answer: B — 2
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