Connecting f, f', and f'' - Calculus 1

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)?
Answer: It is increasing and concave up. 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.
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?
Answer: x = -1 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.
Let $f$ be a function such that $f(x) > 0$ for all $x$ . Let $g(x) = \ln(f(x))$ . If the graph of $g(x)$ is increasing and concave up, which of the following must be true about the graph of $f(x)$ ?
Answer: $f(x)$ is increasing and concave up. 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 $g(x) = \ln(f(x))$ , we can find its derivatives using the chain rule: $g'(x) = \frac{f'(x)}{f(x)}$ $g''(x) = \frac{f''(x) \cdot f(x) - [f'(x)]^2}{[f(x)]^2}$ Given that $g(x)$ is increasing, we know $g'(x) > 0$ , which means $\frac{f'(x)}{f(x)} > 0$ . Since $f(x) > 0$ , we must have $f'(x) > 0$ , so $f(x)$ is increasing. Given that $g(x)$ is concave up, we know $g''(x) > 0$ , which means $\frac{f''(x) \cdot f(x) - [f'(x)]^2}{[f(x)]^2} > 0$ . Since the denominator is always positive, the numerator must be positive: $f''(x) \cdot f(x) - [f'(x)]^2 > 0$ . This gives us $f''(x) \cdot f(x) > [f'(x)]^2$ . Since $f(x) > 0$ and $[f'(x)]^2 > 0$ , we must have $f''(x) > 0$ , so $f(x)$ is concave up. Choice A incorrectly suggests concavity cannot be determined. Choice B incorrectly claims $f(x)$ is concave down. Choice C incorrectly states $f(x)$ is decreasing when we proved it's increasing. Choice D correctly identifies that $f(x)$ 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.
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)?
Answer: f(x) has a local minimum at x=0 and is always concave up. 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 $f(x) = \int_0^x g(t) dt$ , the Fundamental Theorem of Calculus tells us that $f'(x) = g(x)$ . This is crucial for analyzing $f(x)$ 's behavior. Given that $g(t)$ is strictly increasing, we know $f''(x) = g'(x) > 0$ everywhere, so $f(x)$ is always concave up. Since $g(0) < 0$ and $g$ is strictly increasing, $g$ starts negative and eventually becomes positive (crossing zero at some point). This means $f'(x) = g(x)$ starts negative at $x = 0$ , decreases $f$ initially, then becomes positive later, causing $f$ to increase. Therefore, $f$ has a local minimum where $g(x) = 0$ . At $x = 0$ , we have $f'(0) = g(0) < 0$ , so $f$ is decreasing there, but since $f$ is always concave up and will eventually increase, there must be a local minimum nearby (specifically where $g(x) = 0$ ). Option A is wrong because $f$ isn't always increasing—it decreases initially. Option B is wrong because $f$ isn't always decreasing—it eventually increases. Option C is wrong because $f$ has a local minimum (not maximum) where $g(x) = 0$ , and $f$ 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.
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?
Answer: y - 12 = 12(x-2) 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.
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?
Answer: f(x) \ge f(1) for all x. This question tests your understanding of the second derivative test and how concavity relates to local extrema. When you see conditions about $f''(x)$ and $f'(x)$ , think about what they tell you about the function's shape and critical points. Since $f''(x) > 0$ for all $x$ , the function is concave up everywhere. This means the graph curves upward like a smile. Combined with $f'(1) = 0$ , we have a critical point at $x = 1$ where the tangent line is horizontal. The second derivative test tells us that when $f'(c) = 0$ and $f''(c) > 0$ , the function has a local minimum at $x = c$ . Since $f'(1) = 0$ and $f''(1) > 0$ , point $(1, f(1))$ is a local minimum. But because the function is concave up everywhere, this local minimum is actually a global minimum. Therefore, $f(x) \geq f(1)$ for all $x$ , making D correct. Looking at the wrong answers: A is false because $f'(1) = 0$ means the function isn't increasing at $x = 1$ (it momentarily has zero slope). B is incorrect because points of inflection occur where $f''(x) = 0$ or is undefined, but $f''(x) > 0$ everywhere. C confuses the second derivative test—when $f''(x) > 0$ , critical points are local minima, not maxima. Remember: $f''(x) > 0$ 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.
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?
Answer: f has a point of inflection at x=c where the concavity changes from down to up. 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.
Let $f$ be a twice-differentiable function on $[0, 4]$ with $f'(1)=0$ and $f'(3)=0$ . Which of the following statements must be true?
Answer: There exists some $c \in (1, 3)$ such that $f''(c)=0$ . We are given that $f$ is twice-differentiable, which means $f'$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$ . We are also given $f'(1)=0$ and $f'(3)=0$ . Since $f'(1) = f'(3)$ , we can apply Rolle's Theorem to the function $f'(x)$ on the interval $[1, 3]$ . Rolle's Theorem states that there must exist at least one number $c$ in the open interval $(1, 3)$ such that $(f')'(c) = 0$ . The derivative of $f'(x)$ is $f''(x)$ . Therefore, there must exist some $c \in (1, 3)$ such that $f''(c)=0$ . Choice A is a stronger statement; while often true, $f''(c)=0$ does not guarantee an inflection point (concavity might not change). Choice B is not necessarily true; Rolle's Theorem applied to $f$ would require $f(1)=f(3)$ to conclude something about $f'$ , 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 $f$ is not a simple polynomial.
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?
Answer: f has a local maximum at x=1. 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.
Let f be a function such that f(x) > 0, f'(x) < 0, and f''(x) > 0 for all x. If g(x) = (f(x))^2, which of the following must be true for g(x)?
Answer: g is decreasing and concave up. We use the chain rule and product rule to find the derivatives of g(x). g'(x) = 2f(x)f'(x). Since f(x) > 0 and f'(x) < 0, their product is negative. Thus, g'(x) < 0, which means g is decreasing. Next, g''(x) = d/dx [2f(x)f'(x)] = 2(f'(x))^2 + 2f(x)f''(x). The term 2(f'(x))^2 is always positive since it is a square. The term 2f(x)f''(x) is also positive since f(x) > 0 and f''(x) > 0. The sum of two positive terms is positive, so g''(x) > 0. This means g is concave up. Therefore, g is decreasing and concave up.
The first derivative of a function $f$ is given by $f'(x) = (x-1)^2(x-3)$ . Which of the following statements about the function $f$ is correct?
Answer: $f$ has a local minimum at $x=3$ and a point of inflection at $x=1$ . First, we analyze the critical points using the first derivative, $f'(x) = (x-1)^2(x-3)$ . The critical points are where $f'(x)=0$ , which are $x=1$ and $x=3$ . Sign analysis for $f'(x)$ : - If $x<1$ , $f'(x) = (+)(-) < 0$ , so $f$ is decreasing. - If $1<x<3$ , $f'(x) = (+)(-) < 0$ , so $f$ is decreasing. - If $x>3$ , $f'(x) = (+)(+) > 0$ , so $f$ is increasing. At $x=1$ , $f'(x)$ does not change sign, so there is no local extremum. At $x=3$ , $f'(x)$ changes from negative to positive, so $f$ has a local minimum. Next, we find the second derivative to check for points of inflection: $f''(x) = \frac{d}{dx}[(x-1)^2(x-3)] = 2(x-1)(x-3) + (x-1)^2(1) = (x-1)[2(x-3) + (x-1)] = (x-1)(3x-7)$ . The potential inflection points are where $f''(x)=0$ , i.e., at $x=1$ and $x=7/3$ . Sign analysis for $f''(x)$ : - If $x<1$ , $f''(x) = (-)(-) > 0$ , so $f$ is concave up. - If $1<x<7/3$ , $f''(x) = (+)(-) < 0$ , so $f$ is concave down. - If $x>7/3$ , $f''(x) = (+)(+) > 0$ , so $f$ is concave up. Since concavity changes at $x=1$ (from up to down), $f$ has a point of inflection at $x=1$ . Concavity also changes at $x=7/3$ . Combining our findings, $f$ has a local minimum at $x=3$ and a point of inflection at $x=1$ .
Let $f$ be a twice-differentiable function. If $f'(c) = 0$ and $f''(c) < 0$ , which of the following statements must be true about the function $g(x) = e^{f(x)}$ at the point $x=c$ ?
Answer: $g(x)$ has a local maximum at $x=c$ . To determine the behavior of $g(x)$ at $x=c$ , we use the first and second derivative tests. First, we find the derivatives of $g(x)$ using the chain rule. $g'(x) = e^{f(x)} \cdot f'(x)$ $g''(x) = e^{f(x)} \cdot f'(x) \cdot f'(x) + e^{f(x)} \cdot f''(x) = e^{f(x)}[(f'(x))^2 + f''(x)]$ Now, we evaluate these derivatives at $x=c$ using the given information $f'(c) = 0$ and $f''(c) < 0$ . $g'(c) = e^{f(c)} \cdot f'(c) = e^{f(c)} \cdot 0 = 0$ . Since $g'(c) = 0$ , $x=c$ is a critical point of $g(x)$ . $g''(c) = e^{f(c)}[(f'(c))^2 + f''(c)] = e^{f(c)}[0^2 + f''(c)] = e^{f(c)}f''(c)$ . Since $e^{f(c)}$ is always positive and we are given that $f''(c) < 0$ , the product $e^{f(c)}f''(c)$ must be negative. So, $g''(c) < 0$ . By the Second Derivative Test, since $g'(c) = 0$ and $g''(c) < 0$ , the function $g(x)$ has a local maximum at $x=c$ .
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?
Answer: g(x) is non-decreasing for all x. 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.
Let f(x) = x^4 - 4x^3 + 10. On which of the following intervals is the graph of f both decreasing and concave up?
Answer: (2, 3) 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).
A function f is twice-differentiable with f''(x) = x^2(x-5). Which of the following statements about f is true?
Answer: f' has a local minimum at x=5 and f has no point of inflection at x=0. Points of inflection for f occur where f'' changes sign. The roots of f'' are x=0 and x=5. The factor x^2 does not change sign at x=0, so f does not have an inflection point at x=0. The factor (x-5) changes sign from negative to positive at x=5, so f has an inflection point there. The derivative of f' is f''. To find local extrema of f', we look at the sign changes of its derivative, f''. Since f'' changes sign from negative to positive at x=5, f' has a local minimum at x=5. Combining these, statement B is correct.
Which of the following statements is a sufficient condition for a twice-differentiable function f to have a local minimum at x=c?
Answer: f'(c) = 0 and f''(c) > 0 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.
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?
Answer: The graph is above the x-axis, decreasing, and concave down. 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''.
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?
Answer: The function f has an absolute maximum on (a, b) at x=c. 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.
For a twice-differentiable function f, it is known that f(2)=5, f'(2)=0, and f''(x) > 0 for all x in the interval (0, 4). Which of the following could be a value for f(3)?
Answer: 5.5 When you see a problem involving a function's value, its derivative, and second derivative, you're working with concepts that help predict the function's behavior. The key insight here is understanding what each piece of information tells you about how the function behaves near the given point. Given that $f(2) = 5$ , $f'(2) = 0$ , and $f''(x) > 0$ on the interval $(0,4)$ , you can analyze what happens as you move from $x = 2$ to $x = 3$ . Since $f'(2) = 0$ , the function has a horizontal tangent line at $x = 2$ . More importantly, since $f''(x) > 0$ throughout the interval, the function is concave up everywhere, which means $x = 2$ represents a local minimum. Because the function has a local minimum at $x = 2$ where $f(2) = 5$ , and the function is concave up throughout the interval, $f(x)$ must increase as you move away from $x = 2$ in either direction. Therefore, $f(3) > f(2) = 5$ . Looking at the choices: A) 4.8 is impossible because it's less than 5, contradicting our analysis that the function increases from its minimum. B) 5.0 would only occur if the function remained constant from $x = 2$ to $x = 3$ , but positive concavity prevents this. C) is incorrect because we have sufficient information to determine the function's behavior. D) 5.5 is the only value greater than 5, making it consistent with our analysis. Remember: when $f''(x) > 0$ and $f'(a) = 0$ , point $a$ is a local minimum, so the function value must increase as you move away from that point.
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?
Answer: It is decreasing and concave up. 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.

Start this quiz