Q. If f(x) = x^2 + 2x + 1 for x < 0 and f(x) = kx + 1 for x >= 0, find k such that f is differentiable at x = 0.
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Solution
Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.
Correct Answer: A — -1
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Q. If f(x) = x^2 + 2x + 1, find f'(1).
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Solution
f'(x) = 2x + 2, thus f'(1) = 2(1) + 2 = 4.
Correct Answer: C — 3
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Q. If f(x) = x^2 + 2x + 1, what is f'(1)?
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Solution
Calculating the derivative f'(x) = 2x + 2, we find f'(1) = 4.
Correct Answer: B — 3
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Q. If f(x) = x^2 + 2x + 1, what is the vertex of the parabola?
A.
(-1, 0)
B.
(0, 1)
C.
(-1, 1)
D.
(1, 0)
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Solution
The vertex form is f(x) = (x + 1)^2, so the vertex is (-1, 0).
Correct Answer: C — (-1, 1)
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Q. If f(x) = x^2 + 2x + 3, find f'(1).
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Solution
f'(x) = 2x + 2. Therefore, f'(1) = 2(1) + 2 = 4.
Correct Answer: C — 4
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Q. If f(x) = x^2 - 4, what are the x-intercepts?
A.
-2, 2
B.
0, 4
C.
2, 4
D.
None
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Solution
To find x-intercepts, set f(x) = 0: x^2 - 4 = 0, which gives x = ±2.
Correct Answer: A — -2, 2
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Q. If f(x) = x^2 - 4, what is the limit of f(x) as x approaches 2?
A.
0
B.
2
C.
4
D.
Undefined
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Solution
The limit as x approaches 2 is f(2) = 2^2 - 4 = 0.
Correct Answer: C — 4
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Q. If f(x) = x^2 - 4x + 3, what is the value of f(2)?
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Solution
f(2) = 2^2 - 4*2 + 3 = 4 - 8 + 3 = -1.
Correct Answer: A — 0
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Q. If f(x) = x^2 - 4x + 4, find f'(2).
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Solution
f'(x) = 2x - 4. Thus, f'(2) = 2(2) - 4 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^2 and g(x) = x + 1, what is (f ∘ g)(2)?
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Solution
(f ∘ g)(2) = f(g(2)) = f(2 + 1) = f(3) = 3^2 = 9.
Correct Answer: B — 9
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Q. If f(x) = x^2 for x < 1 and f(x) = 2x - 1 for x ≥ 1, is f differentiable at x = 1?
A.
Yes
B.
No
C.
Only continuous
D.
Only left differentiable
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Solution
f'(1) from left = 2 and from right = 2; hence f is differentiable at x = 1.
Correct Answer: B — No
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Q. If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, is f differentiable at x = 0?
A.
Yes
B.
No
C.
Only left differentiable
D.
Only right differentiable
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Solution
Using the limit definition of the derivative, f'(0) exists, hence f is differentiable at x = 0.
Correct Answer: A — Yes
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Q. If f(x) = x^2, what is f(-3)?
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Solution
f(-3) = (-3)^2 = 9.
Correct Answer: C — 9
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Q. If f(x) = x^3 - 3x + 2, find f'(1).
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Solution
f'(x) = 3x^2 - 3. Thus, f'(1) = 3(1)^2 - 3 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^3 - 3x + 2, find the critical points where f'(x) = 0.
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Solution
Set f'(x) = 3x^2 - 3 = 0 and solve for x.
Correct Answer: B — 0
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Q. If f(x) = x^3 - 3x + 2, find the points where f is not differentiable.
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Solution
The function is a polynomial and is differentiable everywhere, hence no points of non-differentiability.
Correct Answer: A — 0
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Q. If f(x) = x^3 - 3x + 2, then f(x) is continuous at:
A.
All x
B.
x = 0
C.
x = 1
D.
x = -1
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Solution
f(x) is a polynomial function and is continuous for all x.
Correct Answer: A — All x
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Q. If f(x) = x^3 - 3x + 2, what is f(1)?
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Solution
f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^3 - 3x + 2, what is the value of f(1)?
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Solution
f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^3 - 3x^2 + 4, find the critical points of f.
A.
x = 0, 1, 2
B.
x = 1, 2
C.
x = 0, 2
D.
x = 1
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Solution
f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives 3x(x - 2) = 0, so x = 0 and x = 2 are critical points.
Correct Answer: B — x = 1, 2
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Q. If f(x) = x^3 - 3x^2 + 4, find the point where f is not differentiable.
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Solution
The function is a polynomial and is differentiable everywhere, but checking critical points shows f'(x) = 0 at x = 2.
Correct Answer: C — 2
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Q. If f(x) = x^3 - 3x^2 + 4, find the point where the function has a local minimum.
A.
(1, 2)
B.
(2, 1)
C.
(3, 4)
D.
(0, 4)
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Solution
f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, f(2)) = (2, 1) is a local minimum.
Correct Answer: A — (1, 2)
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Q. If f(x) = x^3 - 3x^2 + 4, then f'(1) is equal to?
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Solution
f'(x) = 3x^2 - 6x; f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3.
Correct Answer: B — 2
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Q. If f(x) = x^3 - 3x^2 + 4, then f'(2) is equal to?
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Solution
f'(x) = 3x^2 - 6x; f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.
Correct Answer: B — 1
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Q. If f(x) = x^3 - 3x^2 + 4, then the local maxima and minima occur at which of the following points?
A.
(0, 4)
B.
(1, 2)
C.
(2, 2)
D.
(3, 4)
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Solution
To find local maxima and minima, we first find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. Evaluating f(1) = 2 shows it is a local minimum.
Correct Answer: B — (1, 2)
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Q. If f(x) = x^3 - 3x^2 + 4, then the local maxima occurs at which point?
A.
x = 0
B.
x = 1
C.
x = 2
D.
x = 3
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Solution
To find local maxima, we first find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. Checking the second derivative f''(x) = 6x - 6, we find f''(2) < 0, indicating a local maxima at x = 2.
Correct Answer: B — x = 1
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Q. If f(x) = x^3 - 3x^2 + 4, then the local maxima occurs at x = ?
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Solution
To find local maxima, we first find f'(x) = 3x^2 - 6. Setting f'(x) = 0 gives x^2 - 2 = 0, so x = ±√2. Evaluating f''(x) at x = 1 gives f''(1) = 0, indicating a point of inflection. Thus, local maxima occurs at x = 1.
Correct Answer: B — 1
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Q. If f(x) = x^3 - 6x^2 + 9x, find the critical points.
A.
(0, 0)
B.
(3, 0)
C.
(2, 0)
D.
(1, 0)
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Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.
Correct Answer: B — (3, 0)
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Q. If f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1, find f'(1).
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Solution
Calculating f'(x) = 4x^3 - 12x^2 + 12x - 4. Thus, f'(1) = 4 - 12 + 12 - 4 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1, find f'(2).
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Solution
f'(x) = 4x^3 - 12x^2 + 12x - 4; f'(2) = 0.
Correct Answer: A — 0
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