Calculus 1 Quiz: Connecting F F And F
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Connecting F F And FQuestion 1 of 20

Suppose f is a twice-differentiable function such that f''(x) > 0 for all real numbers x, and f'(1)=0. Which of the following statements must be true?

f is an increasing function for all x.
f has a point of inflection at x=1.
f has a local maximum at x=1.
f(x) \ge f(1) for all x.
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Calculus 1 Quiz

Calculus 1 Quiz: Connecting F F And F

Practice Connecting F F And F in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Connecting F F And F, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Suppose f is a twice-differentiable function such that f''(x) > 0 for all real numbers x, and f'(1)=0. Which of the following statements must be true?

  1. f is an increasing function for all x.
  2. f has a point of inflection at x=1.
  3. f has a local maximum at x=1.
  4. f(x) \ge f(1) for all x. (correct answer)
Explanation: This question tests your understanding of the second derivative test and how concavity relates to local extrema. When you see conditions about f′′(x)f''(x) and f′(x)f'(x), think about what they tell you about the function's shape and critical points. Since f′′(x)>0f''(x) > 0 for all xx, the function is concave up everywhere. This means the graph curves upward like a smile. Combined with f′(1)=0f'(1) = 0, we have a critical point at x=1x = 1 where the tangent line is horizontal. The second derivative test tells us that when f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, the function has a local minimum at x=cx = c. Since f′(1)=0f'(1) = 0 and f′′(1)>0f''(1) > 0, point (1,f(1))(1, f(1)) is a local minimum. But because the function is concave up everywhere, this local minimum is actually a global minimum. Therefore, f(x)≥f(1)f(x) \geq f(1) for all xx, making D correct. Looking at the wrong answers: A is false because f′(1)=0f'(1) = 0 means the function isn't increasing at x=1x = 1 (it momentarily has zero slope). B is incorrect because points of inflection occur where f′′(x)=0f''(x) = 0 or is undefined, but f′′(x)>0f''(x) > 0 everywhere. C confuses the second derivative test—when f′′(x)>0f''(x) > 0, critical points are local minima, not maxima. Remember: f′′(x)>0f''(x) > 0 means concave up, and critical points in concave up regions are local minima. Always check what the second derivative tells you about the nature of critical points.

Question 2

On a certain interval I, a function f satisfies f(x) > 0, f'(x) < 0, and f''(x) < 0. Which of the following statements best describes the graph of f on the interval I?

  1. The graph is above the x-axis, increasing, and concave up.
  2. The graph is below the x-axis, decreasing, and concave down.
  3. The graph is above the x-axis, decreasing, and concave down. (correct answer)
  4. The graph is above the x-axis, increasing, and concave down.
Explanation: f(x) > 0 means the graph is above the x-axis. f'(x) < 0 means the function is decreasing. f''(x) < 0 means the graph is concave down. Combining these, the graph is above the x-axis, decreasing, and concave down. Distractors involve misinterpreting the meaning of the signs of f, f', or f''.

Question 3

Let f be a twice-differentiable function on the interval (a, b). If f''(x) < 0 for all x in (a, b) and there exists a c in (a, b) such that f'(c) = 0, which of the following statements must be true?

  1. The function f has a point of inflection at x=c.
  2. The function f has a local minimum at x=c.
  3. The function f has an absolute maximum on (a, b) at x=c. (correct answer)
  4. The function f is decreasing for all x in (a, b).
Explanation: The condition f'(c) = 0 indicates that x=c is a critical point of f. The condition f''(x) < 0 for all x in the interval means the function is concave down everywhere on the interval. By the Second Derivative Test, since f''(c) < 0, the function f has a local maximum at x=c. Because the function is always concave down, there can be only one critical point, and this local maximum must also be the absolute maximum on the interval.

Question 4

Which of the following statements is a sufficient condition for a twice-differentiable function f to have a local minimum at x=c?

  1. f'(c) = 0
  2. f''(c) > 0
  3. f'(c) = 0 and f''(c) > 0 (correct answer)
  4. f'(c) = 0 and f''(c) = 0
Explanation: This question tests the Second Derivative Test. A sufficient condition is a set of criteria that, if met, guarantees the conclusion. f'(c)=0 is a necessary condition for a differentiable function to have a local extremum, but not sufficient (e.g., f(x)=x^3 at c=0). f''(c)>0 only tells us the function is concave up at c, but says nothing about the slope. f'(c)=0 and f''(c)=0 is the inconclusive case. The combination f'(c)=0 (a horizontal tangent) and f''(c)>0 (concave up) is a sufficient condition to guarantee a local minimum.

Question 5

Let ff be a twice-differentiable function. Selected values of ff and its derivatives are given in the table below.

xxf(x)f(x)f′(x)f'(x)f′′(x)f''(x)
051-2
1703
24-31
31-1-4

Let ff be a twice-differentiable function. Selected values of ff and its derivatives are given in the table above. Which of the following statements must be true?

  1. ff has a local minimum at x=1x=1. (correct answer)
  2. The graph of ff has a point of inflection at x=2x=2.
  3. The function ff is decreasing on the interval (1,3)(1, 3).
  4. The graph of f′f' is concave up at x=0x=0.
Explanation: We can use the Second Derivative Test to check for a local minimum at x=1x=1. The table shows that f′(1)=0f'(1) = 0, which means x=1x=1 is a critical point of ff. The table also shows that f′′(1)=3f''(1) = 3. Since f′′(1)>0f''(1) > 0, the function ff is concave up at this critical point. Therefore, ff has a local minimum at x=1x=1. Distractor B is incorrect because for a point of inflection, f′′f'' must change sign, and we don't have enough information to conclude this. Distractor C is incorrect because we cannot guarantee f′(x)f'(x) is negative for the entire interval (1,3)(1, 3); for example, since f′(2)=−3f'(2)=-3 and f′(3)=−1f'(3)=-1, f′f' is increasing on part of this interval. Distractor D is incorrect because the concavity of f′f' is determined by the sign of its derivative, f′′f''. The concavity of ff is determined by f′′f''. At x=0x=0, f′′(0)=−2f''(0) = -2, which means the graph of ff is concave down at x=0x=0, not that the graph of f′f' is concave up.

Question 6

Let h(x) = f(g(x)), where f and g are twice-differentiable functions. Given g(1)=2, g'(1)=-1, g''(1)=3, f'(2)=4, and f''(2)=-2, what can be concluded about the graph of h at x=1?

  1. It is increasing and concave up.
  2. It is increasing and concave down.
  3. It is decreasing and concave up. (correct answer)
  4. It is decreasing and concave down.
Explanation: We must find the signs of h'(1) and h''(1). Using the chain rule: h'(x) = f'(g(x))g'(x). So, h'(1) = f'(g(1))g'(1) = f'(2)(-1) = 4(-1) = -4. Since h'(1) < 0, h is decreasing at x=1. For concavity, we find h''(x) using the product and chain rules: h''(x) = f''(g(x))[g'(x)]^2 + f'(g(x))g''(x). So, h''(1) = f''(g(1))[g'(1)]^2 + f'(g(1))g''(1) = f''(2)(-1)^2 + f'(2)(3) = (-2)(1) + (4)(3) = -2 + 12 = 10. Since h''(1) > 0, h is concave up at x=1. Therefore, h is decreasing and concave up at x=1.

Question 7

Let f(x) = x^4 - 4x^3 + 10. On which of the following intervals is the graph of f both decreasing and concave up?

  1. (-$\infty$, 0)
  2. (0, 2)
  3. (2, 3) (correct answer)
  4. (3, $\infty$)
Explanation: First, find f'(x) and f''(x). f'(x) = 4x^3 - 12x^2 = 4x^2(x-3). f''(x) = 12x^2 - 24x = 12x(x-2). The function f is decreasing when f'(x) < 0, which occurs for x < 3 (and x $\neq$ 0). The function f is concave up when f''(x) > 0, which occurs when x < 0 or x > 2. We need the intersection of these conditions, i.e., where f is both decreasing and concave up. The interval where x<3 and (x<0 or x>2) is (-$\infty$, 0) \cup (2, 3). The only choice that fits is (2, 3).

Question 8

Let f be a twice-differentiable function. If f'(x) > 0 for all x and f(0) = 0, which of the following statements about g(x) = f($x^3$) must be true?

  1. g(x) has a local minimum at x=0.
  2. g(x) has a local maximum at x=0.
  3. g(x) has a point of inflection at x=0.
  4. g(x) is non-decreasing for all x. (correct answer)
Explanation: We analyze the derivative of g(x). Using the chain rule, g'(x) = f'($x^3$) \cdot 3x^2. We are given that f'(y) > 0 for any input y. Therefore, f'($x^3$) is always positive. The term 3x^2 is always non-negative (\geq 0). The product of a positive term and a non-negative term is non-negative. So, g'(x) \geq 0 for all x. A function whose derivative is always non-negative is a non-decreasing function. At x=0, g'(0)=0, but g'(x) is positive on both sides of 0, so there is no local extremum at x=0.

Question 9

Let f be a function such that f(2)=3, f'(2)=0, and f''(2)=-1. What is an equation of the tangent line to the graph of g(x) = x^2 f(x) at x=2?

  1. y - 12 = -4(x-2)
  2. y - 12 = 12(x-2) (correct answer)
  3. y - 6 = 3(x-2)
  4. y - 12 = 0
Explanation: To find the equation of the tangent line, we need a point and a slope. The point is (2, g(2)). g(2) = (2)^2 f(2) = 4 \cdot 3 = 12. The point is (2, 12). The slope is g'(2). We find g'(x) using the product rule: g'(x) = 2x f(x) + x^2 f'(x). Now we evaluate at x=2: g'(2) = 2(2)f(2) + (2)^2 f'(2) = 4(3) + 4(0) = 12. The slope is 12. The equation of the line is y - 12 = 12(x-2). The information f''(2)=-1 is extra information that indicates f has a local maximum at x=2 but is not needed for the tangent line calculation.

Question 10

Let f be a twice-differentiable function and let g(x) = f'(x). If g(x) has a local minimum at x=c, which of the following must be true about the function f?

  1. f has a local minimum at x=c.
  2. f has a local maximum at x=c.
  3. f has a point of inflection at x=c where the concavity changes from down to up. (correct answer)
  4. f has a point of inflection at x=c where the concavity changes from up to down.
Explanation: If g(x) has a local minimum at x=c, then g'(c) = 0 and g'(x) must change sign from negative to positive at x=c. Since g(x) = f'(x), we have g'(x) = f''(x). Therefore, f''(c) = 0 and f''(x) changes sign from negative to positive at x=c. A point where f'' changes sign is a point of inflection. Since the sign of f'' changes from negative to positive, the concavity of f changes from down to up.

Question 11

For a function f at x=a, it is known that f'(a) < 0 and f''(a) > 0. Which statement best describes the function's behavior at x=a?

  1. The function is decreasing and its rate of decrease is slowing down. (correct answer)
  2. The function is decreasing and its rate of decrease is speeding up.
  3. The function is increasing and its rate of increase is slowing down.
  4. The function is increasing and its rate of increase is speeding up.
Explanation: f'(a) < 0 means the function f is decreasing at x=a. The second derivative, f''(a), represents the rate of change of the first derivative, f'(a). Since f''(a) > 0, the slope f'(a) is increasing. As a negative slope increases, it becomes less negative (closer to zero). Therefore, the function is decreasing at a slower rate.

Question 12

The first derivative of a function f is given by f'(x) = (x-4)^2(x+1). At which value of x does f have a local minimum?

  1. x = -1 (correct answer)
  2. x = 4/3
  3. x = 4
  4. f has no local minimum.
Explanation: Local extrema can occur at critical points, where f'(x) = 0. The critical points are x = 4 and x = -1. To determine if they are minima, maxima, or neither, we use the First Derivative Test. We check the sign of f'(x) around these points. The term (x-4)^2 is always non-negative, so it does not affect the sign of f'. The sign of f' is determined by the term (x+1). For x < -1, (x+1) is negative, so f'(x) < 0. For x > -1, (x+1) is positive, so f'(x) > 0. Since f' changes from negative to positive at x = -1, f has a local minimum there. At x=4, f' is positive on both sides, so there is no extremum at x=4.

Question 13

Let ff be a twice-differentiable function. If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, which of the following statements must be true about the function g(x)=ef(x)g(x) = e^{f(x)} at the point x=cx=c?

  1. g(x)g(x) has a local maximum at x=cx=c. (correct answer)
  2. g(x)g(x) has a local minimum at x=cx=c.
  3. g(x)g(x) has a point of inflection at x=cx=c.
  4. The behavior of g(x)g(x) at x=cx=c cannot be determined from the given information.
Explanation: To determine the behavior of g(x)g(x) at x=cx=c, we use the first and second derivative tests. First, we find the derivatives of g(x)g(x) using the chain rule. g′(x)=ef(x)⋅f′(x)g'(x) = e^{f(x)} \cdot f'(x) g′′(x)=ef(x)⋅f′(x)⋅f′(x)+ef(x)⋅f′′(x)=ef(x)[(f′(x))2+f′′(x)]g''(x) = e^{f(x)} \cdot f'(x) \cdot f'(x) + e^{f(x)} \cdot f''(x) = e^{f(x)}[(f'(x))^2 + f''(x)] Now, we evaluate these derivatives at x=cx=c using the given information f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0. g′(c)=ef(c)⋅f′(c)=ef(c)⋅0=0g'(c) = e^{f(c)} \cdot f'(c) = e^{f(c)} \cdot 0 = 0. Since g′(c)=0g'(c) = 0, x=cx=c is a critical point of g(x)g(x). g′′(c)=ef(c)[(f′(c))2+f′′(c)]=ef(c)[02+f′′(c)]=ef(c)f′′(c)g''(c) = e^{f(c)}[(f'(c))^2 + f''(c)] = e^{f(c)}[0^2 + f''(c)] = e^{f(c)}f''(c). Since ef(c)e^{f(c)} is always positive and we are given that f′′(c)<0f''(c) < 0, the product ef(c)f′′(c)e^{f(c)}f''(c) must be negative. So, g′′(c)<0g''(c) < 0. By the Second Derivative Test, since g′(c)=0g'(c) = 0 and g′′(c)<0g''(c) < 0, the function g(x)g(x) has a local maximum at x=cx=c.

Question 14

Let ff be a function such that f(x)>0f(x) > 0 for all xx. Let g(x)=ln⁡(f(x))g(x) = \ln(f(x)). If the graph of g(x)g(x) is increasing and concave up, which of the following must be true about the graph of f(x)f(x)?

  1. f(x)f(x) is increasing, but its concavity cannot be determined.
  2. f(x)f(x) is increasing and concave down.
  3. f(x)f(x) is decreasing and concave up.
  4. f(x)f(x) is increasing and concave up. (correct answer)
Explanation: When you see a question connecting properties of a function and its logarithm, you need to use the chain rule and understand how derivatives relate to increasing/decreasing behavior and concavity. Since g(x)=ln⁡(f(x))g(x) = \ln(f(x)), we can find its derivatives using the chain rule: g′(x)=f′(x)f(x)g'(x) = \frac{f'(x)}{f(x)} g′′(x)=f′′(x)⋅f(x)−[f′(x)]2[f(x)]2g''(x) = \frac{f''(x) \cdot f(x) - [f'(x)]^2}{[f(x)]^2} Given that g(x)g(x) is increasing, we know g′(x)>0g'(x) > 0, which means f′(x)f(x)>0\frac{f'(x)}{f(x)} > 0. Since f(x)>0f(x) > 0, we must have f′(x)>0f'(x) > 0, so f(x)f(x) is increasing. Given that g(x)g(x) is concave up, we know g′′(x)>0g''(x) > 0, which means f′′(x)⋅f(x)−[f′(x)]2[f(x)]2>0\frac{f''(x) \cdot f(x) - [f'(x)]^2}{[f(x)]^2} > 0. Since the denominator is always positive, the numerator must be positive: f′′(x)⋅f(x)−[f′(x)]2>0f''(x) \cdot f(x) - [f'(x)]^2 > 0. This gives us f′′(x)⋅f(x)>[f′(x)]2f''(x) \cdot f(x) > [f'(x)]^2. Since f(x)>0f(x) > 0 and [f′(x)]2>0[f'(x)]^2 > 0, we must have f′′(x)>0f''(x) > 0, so f(x)f(x) is concave up. Choice A incorrectly suggests concavity cannot be determined. Choice B incorrectly claims f(x)f(x) is concave down. Choice C incorrectly states f(x)f(x) is decreasing when we proved it's increasing. Choice D correctly identifies that f(x)f(x) is both increasing and concave up. Remember: when working with composite functions involving logarithms, carefully apply the chain rule and use the signs of derivatives to determine monotonicity and concavity.

Question 15

The first derivative of a function ff is given by f′(x)=(x−1)2(x−3)f'(x) = (x-1)^2(x-3). Which of the following statements about the function ff is correct?

  1. ff has a local minimum at x=1x=1 and a local maximum at x=3x=3.
  2. ff has a local minimum at x=3x=3 and a point of inflection at x=1x=1. (correct answer)
  3. ff has a local maximum at x=1x=1 and a point of inflection at x=3x=3.
  4. ff has points of inflection at x=1x=1 and x=3x=3.
Explanation: First, we analyze the critical points using the first derivative, f′(x)=(x−1)2(x−3)f'(x) = (x-1)^2(x-3). The critical points are where f′(x)=0f'(x)=0, which are x=1x=1 and x=3x=3. Sign analysis for f′(x)f'(x):
  • If x<1x<1, f′(x)=(+)(−)<0f'(x) = (+)(-) < 0, so ff is decreasing.
  • If 1<x<31<x<3, f′(x)=(+)(−)<0f'(x) = (+)(-) < 0, so ff is decreasing.
  • If x>3x>3, f′(x)=(+)(+)>0f'(x) = (+)(+) > 0, so ff is increasing. At x=1x=1, f′(x)f'(x) does not change sign, so there is no local extremum. At x=3x=3, f′(x)f'(x) changes from negative to positive, so ff has a local minimum. Next, we find the second derivative to check for points of inflection: f′′(x)=ddx[(x−1)2(x−3)]=2(x−1)(x−3)+(x−1)2(1)=(x−1)[2(x−3)+(x−1)]=(x−1)(3x−7)f''(x) = \frac{d}{dx}[(x-1)^2(x-3)] = 2(x-1)(x-3) + (x-1)^2(1) = (x-1)[2(x-3) + (x-1)] = (x-1)(3x-7). The potential inflection points are where f′′(x)=0f''(x)=0, i.e., at x=1x=1 and x=7/3x=7/3. Sign analysis for f′′(x)f''(x):
  • If x<1x<1, f′′(x)=(−)(−)>0f''(x) = (-)(-) > 0, so ff is concave up.
  • If 1<x<7/31<x<7/3, f′′(x)=(+)(−)<0f''(x) = (+)(-) < 0, so ff is concave down.
  • If x>7/3x>7/3, f′′(x)=(+)(+)>0f''(x) = (+)(+) > 0, so ff is concave up. Since concavity changes at x=1x=1 (from up to down), ff has a point of inflection at x=1x=1. Concavity also changes at x=7/3x=7/3. Combining our findings, ff has a local minimum at x=3x=3 and a point of inflection at x=1x=1.

Question 16

Let ff be a twice-differentiable function on [0,4][0, 4] with f′(1)=0f'(1)=0 and f′(3)=0f'(3)=0. Which of the following statements must be true?

  1. ff has at least one point of inflection in the interval (1,3)(1, 3).
  2. f(1)=f(3)f(1) = f(3).
  3. There exists some c∈(1,3)c \in (1, 3) such that f′′(c)=0f''(c)=0. (correct answer)
  4. ff has a local maximum at x=1x=1 and a local minimum at x=3x=3.
Explanation: We are given that ff is twice-differentiable, which means f′f' is continuous on [1,3][1, 3] and differentiable on (1,3)(1, 3). We are also given f′(1)=0f'(1)=0 and f′(3)=0f'(3)=0. Since f′(1)=f′(3)f'(1) = f'(3), we can apply Rolle's Theorem to the function f′(x)f'(x) on the interval [1,3][1, 3]. Rolle's Theorem states that there must exist at least one number cc in the open interval (1,3)(1, 3) such that (f′)′(c)=0(f')'(c) = 0. The derivative of f′(x)f'(x) is f′′(x)f''(x). Therefore, there must exist some c∈(1,3)c \in (1, 3) such that f′′(c)=0f''(c)=0. Choice A is a stronger statement; while often true, f′′(c)=0f''(c)=0 does not guarantee an inflection point (concavity might not change). Choice B is not necessarily true; Rolle's Theorem applied to ff would require f(1)=f(3)f(1)=f(3) to conclude something about f′f', not the other way around. Choice D is not necessary; the order of the maximum and minimum could be reversed, or they could both be of the same type if ff is not a simple polynomial.

Question 17

For a function f, its derivative f'(x) is a continuous and strictly decreasing function for all real numbers. If f'(1)=0, which of the following statements must be true about f?

  1. f has a local minimum at x=1.
  2. f has a local maximum at x=1. (correct answer)
  3. f has a point of inflection at x=1.
  4. f is a decreasing function for all x.
Explanation: The fact that f'(x) is strictly decreasing means its derivative, f''(x), must be less than or equal to zero. Given f'(x) is continuous, f''(x) < 0 for almost all x. Since f'(x) is strictly decreasing and f'(1)=0, it must be that f'(x) > 0 for x<1 and f'(x) < 0 for x>1. According to the First Derivative Test, since f' changes from positive to negative at x=1, the function f has a local maximum at x=1. This is also consistent with the Second Derivative Test, as f''(1) would be negative.

Question 18

Let f be a function such that f'(x) = g(x). If g(x) is a strictly positive and strictly increasing function for all x, which statement accurately describes the graph of f(x)?

  1. It is increasing and concave up. (correct answer)
  2. It is increasing and concave down.
  3. It is decreasing and concave up.
  4. It is decreasing and concave down.
Explanation: The derivative of f is f'(x)=g(x). Since g(x) is strictly positive, f'(x) > 0 for all x, which means f(x) is a strictly increasing function. The second derivative of f is f''(x)=g'(x). Since g(x) is a strictly increasing function, its derivative g'(x) must be positive. Therefore, f''(x) > 0 for all x, which means f(x) is concave up. Combining these two facts, the graph of f is increasing and concave up.

Question 19

Let the first derivative of a function f be defined as f'(x) = |x-2|. Which of the following statements about f is true?

  1. f has a local minimum at x=2.
  2. f has a point of inflection at x=2. (correct answer)
  3. f is concave up for all x \neq 2.
  4. f has a local maximum at x=2.
Explanation: f'(x) = |x-2| is 0 at x=2 and positive for all x \neq 2. Since f'(x) does not change sign at x=2, f does not have a local extremum there. To analyze concavity, we look at f''(x). For x>2, f'(x)=x-2, so f''(x)=1. For x<2, f'(x)=-(x-2)=2-x, so f''(x)=-1. Since f''(x) changes sign from negative to positive at x=2, the concavity of f changes from down to up. Therefore, f has a point of inflection at x=2.

Question 20

A function f is defined by f(x) = \int_0^x g(t) dt. If g(t) is a continuous and strictly increasing function for all t, and g(0) < 0, which statement must be true about f(x)?

  1. f(x) is always increasing and always concave up.
  2. f(x) is always decreasing and always concave up.
  3. f(x) has a local maximum at the x value where g(x)=0 and is always concave down.
  4. f(x) has a local minimum at x=0 and is always concave up. (correct answer)
Explanation: This question tests your understanding of the Fundamental Theorem of Calculus and how properties of the integrand affect the resulting function. When you see a function defined as an integral with a variable upper limit, immediately think about using the FTC to find derivatives. Since f(x)=∫0xg(t)dtf(x) = \int_0^x g(t) dt, the Fundamental Theorem of Calculus tells us that f′(x)=g(x)f'(x) = g(x). This is crucial for analyzing f(x)f(x)'s behavior. Given that g(t)g(t) is strictly increasing, we know f′′(x)=g′(x)>0f''(x) = g'(x) > 0 everywhere, so f(x)f(x) is always concave up. Since g(0)<0g(0) < 0 and gg is strictly increasing, gg starts negative and eventually becomes positive (crossing zero at some point). This means f′(x)=g(x)f'(x) = g(x) starts negative at x=0x = 0, decreases ff initially, then becomes positive later, causing ff to increase. Therefore, ff has a local minimum where g(x)=0g(x) = 0. At x=0x = 0, we have f′(0)=g(0)<0f'(0) = g(0) < 0, so ff is decreasing there, but since ff is always concave up and will eventually increase, there must be a local minimum nearby (specifically where g(x)=0g(x) = 0). Option A is wrong because ff isn't always increasing—it decreases initially. Option B is wrong because ff isn't always decreasing—it eventually increases. Option C is wrong because ff has a local minimum (not maximum) where g(x)=0g(x) = 0, and ff is concave up (not down). Remember: when analyzing functions defined by integrals, immediately apply the FTC to find the derivative, then analyze the sign changes and monotonicity of that derivative.