Graphing Functions & Derivatives - Calculus 1

1. Let $f$ be a twice-differentiable function with $f'(x) > 0$ and $f''(x) < 0$ on an interval $I$. Let $g(x) = e^{-f(x)}$. Which of the following describes the graph of $g(x)$ on $I$?
   Answer: Decreasing and concave up We find the first and second derivatives of $g(x)$. Using the chain rule, $g'(x) = e^{-f(x)} \cdot (-f'(x))$. Since $e^{-f(x)}$ is always positive and we are given $f'(x) > 0$, $g'(x)$ must be negative. Thus, $g(x)$ is decreasing. For the second derivative, we use the product and chain rules: $g''(x) = e^{-f(x)}(-f'(x))^2 + e^{-f(x)}(-f''(x)) = e^{-f(x)}((f'(x))^2 - f''(x))$. We are given $f'(x) > 0$, so $(f'(x))^2 > 0$. We are given $f''(x) < 0$, so $-f''(x) > 0$. The sum of two positive numbers is positive. Since $e^{-f(x)}$ is also positive, $g''(x)$ must be positive. Thus, $g(x)$ is concave up.
2. Let $f$ be a twice-differentiable function. If $f'(2) = 0$ and the graph of $f''(x)$ is a line with a positive slope and an x-intercept at $x=2$, what feature does the graph of $f(x)$ have at $x=2$?
   Answer: A point of inflection We are given $f'(2) = 0$, so $x=2$ is a critical point. The information about $f''(x)$ tells us its properties. An x-intercept at $x=2$ means $f''(2) = 0$. Because $f''(2)=0$, the Second Derivative Test for extrema is inconclusive. A line with positive slope means that for $x<2$, $f''(x)$ is negative, and for $x>2$, $f''(x)$ is positive. Since $f''(x)$ changes sign at $x=2$, the graph of $f(x)$ has a point of inflection at $x=2$. Because the sign of $f'$ does not change (it is decreasing to $f'(2)=0$ and then continues decreasing since $f''$ is negative then positive... wait, this is wrong. Let's re-analyze $f'$. We know $f'(2)=0$ and $f''$ changes from neg to pos. This means $f'$ is decreasing for $x<2$ and increasing for $x>2$. This implies $f'$ has a local minimum at $x=2$. Since $f'(2)=0$, it must be that $f'(x) > 0$ for $x$ near 2. So $f$ is always increasing. Thus no extremum. Let's re-read my analysis. At $x=2$, $f'(2)=0$ and $f''$ changes sign. The First Derivative Test still applies. Since $f'' < 0$ for $x < 2$, $f'$ is decreasing. Since $f'' > 0$ for $x > 2$, $f'$ is increasing. This means $f'$ has a minimum at $x=2$. Since $f'(2)=0$, $f'(x)$ must be positive on both sides of $x=2$ (for $x$ near 2). So $f'$ does not change sign. Therefore, there is no local extremum. There is, however, an inflection point because $f''$ changes sign.
3. Suppose $f$ is a function such that for all $x$ in the interval $(-1, 1)$, $f'(x) < 0$ and $f''(x) > 0$. Which of the following statements must be true about the graph of its derivative, $y=f'(x)$, on $(-1, 1)$?
   Answer: The graph of $f'(x)$ is below the x-axis and is increasing. The condition $f'(x) < 0$ means that the value of the function $f'(x)$ is negative. On a graph, this means the graph of $y=f'(x)$ lies below the x-axis. The condition $f''(x) > 0$ means that the derivative of $f'(x)$ is positive. When the derivative of a function is positive, that function is increasing. Therefore, the function $f'(x)$ is increasing. Combining these two facts, the graph of $y=f'(x)$ is below the x-axis and is increasing.
4. The derivative of a function $f$ is given by $f'(x) = \cos(x) - x$. On the interval $[0, 2\pi]$, the function $f(x)$ is concave down on the set of all $x$ for which...
   Answer: $-\sin(x) < 1$ The graph of $f(x)$ is concave down where $f''(x) < 0$. We find the second derivative: $f''(x) = \frac{d}{dx}(\cos(x) - x) = -\sin(x) - 1$. For concave down behavior, we need $-\sin(x) - 1 < 0$, which simplifies to $-\sin(x) < 1$. Since $-1 \leq \sin(x) \leq 1$, we have $-1 \leq -\sin(x) \leq 1$. Therefore $-\sin(x) < 1$ is satisfied for all $x$ in $[0, 2\pi]$ except at $x = 3\pi/2$ where $\sin(x) = -1$ and $-\sin(x) = 1$.
5. Consider the function $f(x) = x^{2/3}(x-5)$. Which statement correctly describes a key feature of the graph of $f$?
   Answer: The graph has a local maximum at $(0, 0)$ and a local minimum at $(2, -3\sqrt[3]{4})$. The derivative is $f'(x) = \frac{d}{dx}(x^{5/3} - 5x^{2/3}) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5(x-2)}{3x^{1/3}}$. The critical points occur where $f'(x)=0$ (at $x=2$) and where $f'(x)$ is undefined (at $x=0$). Analyzing the sign of $f'$: for $x<0$, $f'>0$; for $0<x<2$, $f'<0$; for $x>2$, $f'>0$. This indicates a local maximum at $x=0$ and a local minimum at $x=2$. The function values are $f(0)=0$ and $f(2)=2^{2/3}(2-5) = -3\sqrt[3]{4}$. As $x \to 0$, $f'(x) \to \pm\infty$, indicating a cusp (a non-differentiable point) at $(0,0)$, not a horizontal tangent. The second derivative is $f''(x) = \frac{10(x+1)}{9x^{4/3}}$, which shows an inflection point at $x=-1$ but not at $x=0$.
6. A function $f$ is twice-differentiable and its first derivative is given by $f'(x) = x^2(x-4)$. Which of the following correctly describes the graph of $y=f(x)$?
   Answer: One local minimum and two points of inflection. Critical points of $f(x)$ occur where $f'(x) = 0$, which are $x=0$ and $x=4$. Analyzing the sign of $f'(x)$, for $x<0$, $f'(x)<0$. For $0<x<4$, $f'(x)<0$. For $x>4$, $f'(x)>0$. Since the sign of $f'(x)$ does not change at $x=0$, there is no extremum there. The sign changes from negative to positive at $x=4$, so there is a local minimum at $x=4$. To find inflection points, we analyze $f''(x)$. $f'(x) = x^3 - 4x^2$, so $f''(x) = 3x^2 - 8x = x(3x-8)$. The potential inflection points are at $x=0$ and $x=8/3$. The sign of $f''(x)$ changes at both of these values (from positive to negative at $x=0$, and from negative to positive at $x=8/3$). Therefore, there are two points of inflection. The function has one local minimum and two points of inflection.
7. The function $f(x)$ is continuous for all real numbers and its derivative is $f'(x) = (x-2)^{-1/3}$. What feature does the graph of $y=f(x)$ have at $x=2$?
   Answer: A vertical tangent The derivative $f'(x) = \frac{1}{\sqrt[3]{x-2}}$ is undefined at $x=2$ because the denominator is zero. As $x \to 2$, the magnitude of $f'(x)$ approaches infinity. This indicates that the slope of the tangent line to the graph of $f(x)$ becomes infinitely steep, which is a vertical tangent. The function is given as continuous, so it cannot have a vertical asymptote or a jump discontinuity. A corner has differing finite slopes from the left and right, which is not the case here.
8. The function $f$ is three times differentiable. A necessary condition for $f$ to have a point of inflection at $x=c$ is $f''(c) = 0$. Which of the following is a sufficient condition?
   Answer: $f''(c)=0$ and $f'''(c) ≠ 0$ A point of inflection requires the concavity to change, which means the sign of $f''(x)$ must change as $x$ passes through $c$. The condition $f''(c) = 0$ alone is not sufficient (e.g., $f(x) = x^4$ at $x=0$). The Third Derivative Test for inflection points states that if $f''(c) = 0$ and $f'''(c) ≠ 0$, then $f''(x)$ must change sign at $x=c$, guaranteeing a point of inflection. If $f'''(c) = 0$, the test is inconclusive.
9. Let $f$ be a differentiable function. The graph of $f$ passes through the points $(0,1)$ and $(2,5)$. It is also known that $f'(x)$ is an increasing function. Which of the following statements must be true?
   Answer: $f(1) < 3$ The condition that $f'(x)$ is increasing means that $f(x)$ is concave up. The secant line connecting the points $(0,1)$ and $(2,5)$ has the equation $y - 1 = \frac{5-1}{2-0}(x-0)$, which simplifies to $y = 2x + 1$. For a concave up function, the graph of the function on the interval $(0,2)$ must lie below its secant line on that interval. At $x=1$, the value on the secant line is $y = 2(1) + 1 = 3$. Therefore, it must be true that $f(1) < 3$.
10. Let $f$ be a twice-differentiable function on $(-\infty, \infty)$. The graph of $f$ is concave up on $(-\infty, 3)$ and concave down on $(3, \infty)$. Which of the following statements must be true about the derivative function $f'$?
    Answer: $f'$ is increasing on $(-\infty, 3)$ and decreasing on $(3, \infty)$. The concavity of $f$ is determined by the sign of $f''$. Concave up means $f''(x) > 0$, and concave down means $f''(x) < 0$. The behavior of $f'$ (whether it is increasing or decreasing) is also determined by the sign of its derivative, $f''$. Therefore, on $(-\infty, 3)$, since $f$ is concave up, $f''(x) > 0$, which implies that $f'$ is increasing. On $(3, \infty)$, since $f$ is concave down, $f''(x) < 0$, which implies that $f'$ is decreasing. This describes the behavior of $f'$ correctly.
11. The function $f(x)$ has a local minimum at $x=c$. Which of the following statements about the graph of $g(x) = -f(x)$ must be true?
    Answer: The graph of $g(x)$ has a local maximum at $x=c$. The graph of $g(x) = -f(x)$ is a reflection of the graph of $f(x)$ across the x-axis. A local minimum is a point that is lower than all nearby points. When this point is reflected across the x-axis, it becomes a point that is higher than all nearby reflected points. Therefore, the graph of $g(x)$ must have a local maximum at $x=c$. Analytically, if $f$ has a local minimum at $c$, then in a neighborhood of $c$, $f(x) \ge f(c)$. Multiplying by -1 reverses the inequality: $-f(x) \le -f(c)$, which means $g(x) \le g(c)$, the definition of a local maximum.
12. Let $f$ be a function such that $f'(x) = (x-a)^2 g(x)$ where $a$ is a constant and $g$ is a differentiable function with $g(a) ≠ 0$. What does the graph of $f$ have at $x=a$?
    Answer: A horizontal tangent but no local extremum. A critical point occurs where $f'(x)=0$, which is at $x=a$. Thus, there is a horizontal tangent at $x=a$. To determine if it is a local extremum, we use the First Derivative Test. The factor $(x-a)^2$ is always non-negative. Since $g(x)$ is continuous and $g(a) ≠ 0$, the sign of $g(x)$ will be the same in a small neighborhood around $x=a$. Therefore, the sign of $f'(x)$ does not change as $x$ passes through $a$. Because the sign of the derivative does not change, $f$ has no local extremum at $x=a$.
13. A function $f$ is defined such that its derivative is $f'(x) = x^2(x-3)$. Which of the following statements accurately describes the graph of $f$?
    Answer: $f$ has a local minimum at $x=3$ and inflection points at $x=0$ and $x=2$. First, we find the critical points by setting $f'(x) = 0$, which gives $x=0$ and $x=3$. Analyzing the sign of $f'(x)$: for $x<0$, $f'(x)<0$; for $0<x<3$, $f'(x)<0$; for $x>3$, $f'(x)>0$. Since $f'$ does not change sign at $x=0$, there is no local extremum there. Since $f'$ changes from negative to positive at $x=3$, there is a local minimum at $x=3$. Next, we find the second derivative: $f''(x) = \frac{d}{dx}(x^3-3x^2) = 3x^2-6x = 3x(x-2)$. Setting $f''(x)=0$ gives $x=0$ and $x=2$. Analyzing the sign of $f''$: for $x<0$, $f''>0$; for $0<x<2$, $f''<0$; for $x>2$, $f''>0$. Since the concavity changes at both $x=0$ and $x=2$, these are the locations of inflection points. Thus, $f$ has a local minimum at $x=3$ and inflection points at $x=0$ and $x=2$.
14. Let $f$ be a differentiable function such that $f(-x) = f(x)$ for all $x$ in its domain. Which of the following statements must be true about its derivative, $f'(x)$?
    Answer: $f'(x)$ is an odd function. The condition $f(-x) = f(x)$ defines an even function. To find the property of its derivative, we differentiate both sides of the equation with respect to $x$. Using the chain rule on the left side, we get $\frac{d}{dx}f(-x) = f'(-x) \cdot (-1) = -f'(-x)$. The derivative of the right side is $\frac{d}{dx}f(x) = f'(x)$. Setting them equal gives $-f'(-x) = f'(x)$, or $f'(-x) = -f'(x)$. This is the definition of an odd function.
15. The graphs of a function $f(x)$ and its derivative $f'(x)$ are related in many ways. If the graph of $f(x)$ has an inflection point at $x=c$, which of the following graphical features corresponds to this on the graph of $f'(x)$?
    Answer: A local maximum or a local minimum at $x=c$. A point of inflection on the graph of $f(x)$ occurs at $x=c$ when the concavity changes. This means that the sign of the second derivative, $f''(x)$, changes at $x=c$. Since $f''(x)$ is the derivative of $f'(x)$, a sign change in $f''(x)$ means that the function $f'(x)$ changes from increasing to decreasing, or vice-versa. A point where a function changes from increasing to decreasing (or vice versa) is a local maximum or a local minimum. Therefore, an inflection point of $f(x)$ corresponds to a local extremum of $f'(x)$.
16. The function $g(x)$ has derivatives of all orders. The graph of its derivative, $g'(x)$, has a local maximum at $x=c$. Which of the following must be true about the graph of $g(x)$ at $x=c$?
    Answer: The graph of $g(x)$ has a point of inflection. If $g'(x)$ has a local maximum at $x=c$, then its derivative must be zero at that point, and the sign of its derivative must change from positive to negative. The derivative of $g'(x)$ is $g''(x)$. So, $g''(c) = 0$ and $g''(x)$ changes sign at $x=c$. A point on the graph of $g(x)$ where $g''(x)$ changes sign is a point of inflection. We cannot conclude anything about local extrema of $g(x)$ because we do not know the value of $g'(c)$.
17. The graph of $f'(x)$, the derivative of a function $f(x)$, is a line with slope -2 that passes through the point $(3, 0)$. Which statement accurately describes the graph of $f(x)$?
    Answer: The graph of $f(x)$ is always concave down and has a local maximum at $x=3$. The graph of $f'(x)$ is a line with slope -2, so $f''(x) = -2$. Since $f''(x) < 0$ for all $x$, the graph of $f(x)$ is always concave down. The line for $f'(x)$ passes through $(3,0)$, meaning $f'(3)=0$, so $x=3$ is a critical point of $f(x)$. Since the slope of $f'(x)$ is negative, for $x<3$, $f'(x)>0$ and for $x>3$, $f'(x)<0$. By the First Derivative Test, since $f'$ changes from positive to negative at $x=3$, $f(x)$ has a local maximum at $x=3$.
18. Let $h(x) = \int_{0}^{x} (t^2 - 3t + 2) \, dt$. On which open interval is the graph of $h(x)$ both decreasing and concave up?
    Answer: ($3/2, 2$) By the Fundamental Theorem of Calculus, Part 1, $h'(x) = x^2 - 3x + 2 = (x-1)(x-2)$. The function $h(x)$ is decreasing when $h'(x) < 0$, which occurs on the interval $(1, 2)$. To determine concavity, we find the second derivative: $h''(x) = 2x - 3$. The graph of $h(x)$ is concave up when $h''(x) > 0$, which means $2x - 3 > 0$, or $x > 3/2$. We need the interval where both conditions are met, which is the intersection of $(1, 2)$ and $(3/2, \infty)$. This intersection is the interval $(3/2, 2)$.
19. Let $f(x) = x^k e^{-x}$, where $k$ is a positive integer. For what values of $k$ does the graph of $f(x)$ have a point of inflection at $x=2+\sqrt{2}$?
    Answer: $k=2$ only We need to find the second derivative and set it to zero. $f'(x) = kx^{k-1}e^{-x} - x^k e^{-x} = e^{-x}(kx^{k-1} - x^k)$. Then $f''(x) = -e^{-x}(kx^{k-1} - x^k) + e^{-x}(k(k-1)x^{k-2} - kx^{k-1}) = e^{-x}[-kx^{k-1} + x^k + k(k-1)x^{k-2} - kx^{k-1}] = x^{k-2}e^{-x}[-kx + x^2 + k(k-1) - kx] = x^{k-2}e^{-x}[x^2 - 2kx + k(k-1)]$. For an inflection point, we set the bracketed quadratic to zero: $x^2 - 2kx + k(k-1) = 0$. We can solve for $x$ using the quadratic formula: $x = \frac{2k \pm \sqrt{4k^2 - 4(k^2-k)}}{2} = k \pm \sqrt{k^2 - (k^2-k)} = k \pm \sqrt{k}$. We are given that an inflection point is at $x=2+\sqrt{2}$. Comparing this with $k \pm \sqrt{k}$, we can see by inspection that $k=2$. For $k=2$, the inflection points are at $x = 2 \pm \sqrt{2}$. Thus, $k=2$ is the correct value.
20. For a function $f$, $f(3) = 5$ and $f'(3) = 0$. Also, $f'(x) > 0$ for $x > 3$ and $f'(x) < 0$ for $x < 3$. Which of the following statements about the graph of $f$ is consistent with this information?
    Answer: The graph has a local minimum at $(3,5)$ and is concave up at $x=3$. The information about $f'(x)$ indicates that it changes from negative to positive at $x=3$. By the First Derivative Test, this means $f$ has a local minimum at $x=3$. The value of the minimum is $f(3)=5$. The information $f'(x) < 0$ for $x<3$ and $f'(x) > 0$ for $x>3$ means that the function $f'(x)$ is increasing as it passes through $x=3$. A function is increasing when its derivative is positive. The derivative of $f'(x)$ is $f''(x)$. Therefore, at $x=3$, we can infer that $f''(3) \ge 0$. A positive second derivative indicates that the graph of $f$ is concave up. Thus, the graph has a local minimum at $(3,5)$ and is concave up at $x=3$.

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