EVT, Extrema & Critical Points - Calculus 1

1. A function is defined by $f(x) = \frac{x^2+3}{x-1}$ on the interval $[2, 5]$. What is the sum of the absolute maximum and absolute minimum values of $f$ on this interval?
   Answer: 13 First, find the derivative: $f'(x) = \frac{(2x)(x-1) - (x^2+3)(1)}{(x-1)^2} = \frac{x^2-2x-3}{(x-1)^2} = \frac{(x-3)(x+1)}{(x-1)^2}$. The only critical point in $[2, 5]$ is $x=3$. The function is continuous on this interval. We check the function values at the critical point and the endpoints: $f(2) = \frac{7}{1} = 7$, $f(3) = \frac{12}{2} = 6$, and $f(5) = \frac{28}{4} = 7$. The absolute minimum value is 6 and the absolute maximum value is 7. Their sum is $6+7=13$.
2. For what values of the constant $k$ does the function $f(x) = x^3 - kx^2 + 3x$ have exactly one critical point?
   Answer: $k=3$ or $k=-3$ Critical points occur where the derivative $f'(x)$ is zero or undefined. Since $f(x)$ is a polynomial, its derivative will be defined everywhere. First, find the derivative: $f'(x) = 3x^2 - 2kx + 3$. To find the critical points, we set the derivative equal to zero: $3x^2 - 2kx + 3 = 0$. This is a quadratic equation in terms of $x$. The number of critical points is the number of real solutions to this equation. The function will have exactly one critical point if this quadratic equation has exactly one real solution. This occurs when the discriminant ($b^2 - 4ac$) is equal to zero. For the quadratic $3x^2 - 2kx + 3 = 0$, we have $a=3$, $b=-2k$, and $c=3$. The discriminant is: $(-2k)^2 - 4(3)(3) = 4k^2 - 36$. Set the discriminant to zero and solve for $k$: $4k^2 - 36 = 0$ $4k^2 = 36$ $k^2 = 9$ $k = \pm 3$ Thus, the function has exactly one critical point when $k=3$ or $k=-3$.
3. Which statement accurately describes the extrema of $f$ on the interval $[-2, 4]$?
   Answer: The absolute maximum is 6 and the absolute minimum is 0. We must check the critical points and the endpoints of the interval $[-2, 4]$. For $x < 1$, $f'(x) = -2x$, so $x=0$ is a critical point. For $x > 1$, $f'(x) = 1$, so there are no critical points. At $x=1$, the function is continuous, but the left-hand derivative is -2 and the right-hand derivative is 1, so $x=1$ is a critical point (a corner). The candidates for absolute extrema are the endpoints $-2, 4$ and the critical points $0, 1$. Evaluating the function at these points: $f(-2)=0$, $f(0)=4$, $f(1)=3$, $f(4)=6$. Comparing these values, the absolute maximum is 6 and the absolute minimum is 0.
4. Consider the function $g(x) = \tan(x)$ on the interval $[0, \pi]$. Which of the following explains why the Extreme Value Theorem does not guarantee that $g(x)$ has an absolute maximum and an absolute minimum on this interval?
   Answer: The function $g(x)$ is not continuous on the interval $[0, \pi]$. The Extreme Value Theorem has two conditions: the function must be continuous and the interval must be closed. The interval $[0, \pi]$ is closed. However, $g(x) = \tan(x)$ has an infinite discontinuity at $x=\pi/2$, which lies within the interval. Because the function is not continuous on the entire interval, the theorem's conditions are not met, and it does not apply.
5. Let $h(x) = \sin(e^x)$. How many critical points does $h(x)$ have on the interval $[0, \ln(2\pi)]$?
   Answer: 2 When you encounter questions about critical points, remember that these occur where the derivative equals zero or is undefined. For composite functions like this one, you'll need the chain rule to find where the derivative vanishes. To find critical points of $h(x) = \sin(e^x)$, start by taking the derivative: $h'(x) = \cos(e^x) \cdot e^x$. Critical points occur when $h'(x) = 0$, which happens when $\cos(e^x) = 0$ (since $e^x$ is never zero). The cosine function equals zero when its argument is an odd multiple of $\frac{\pi}{2}$. So we need: $e^x = \frac{\pi}{2} + n\pi = \frac{(2n+1)\pi}{2}$ where $n$ is any integer. Taking the natural logarithm: $x = \ln\left(\frac{(2n+1)\pi}{2}\right)$ Now check which values fall in $[0, \ln(2\pi)]$: - For $n = 0$: $x = \ln\left(\frac{\pi}{2}\right) \approx 0.45$, which is in our interval - For $n = 1$: $x = \ln\left(\frac{3\pi}{2}\right) \approx 1.50$, which is also in our interval since $\ln(2\pi) \approx 1.84$ - For $n = 2$: $x = \ln\left(\frac{5\pi}{2}\right) \approx 2.25$, which exceeds $\ln(2\pi)$ Therefore, there are exactly 2 critical points. Choice A) suggests only 1 critical point, missing the second one. Choice C) incorrectly counts 3, perhaps including an endpoint or miscalculating the interval bounds. Choice B) confuses this with a function having infinitely many critical points over all real numbers, not recognizing the interval constraint. Strategy tip: For composite trig functions, always solve the inner function equation systematically and carefully check interval boundaries to count critical points accurately.
6. For what range of values of the constant $k$ does the function $f(x) = x^3 - kx^2 + 27x - 1$ have no local extrema?
   Answer: $-9 \leq k \leq 9$ A function has no local extrema if its derivative never changes sign. The derivative is $f'(x) = 3x^2 - 2kx + 27$. This is a quadratic function. For $f'(x)$ to not change sign, it must have at most one real root. This occurs when the discriminant of the quadratic is less than or equal to zero. The discriminant is $\Delta = (-2k)^2 - 4(3)(27) = 4k^2 - 324$. We need $4k^2 - 324 \leq 0$, which simplifies to $k^2 \leq 81$. This inequality is true for $-9 \leq k \leq 9$.
7. A function $f$ is twice-differentiable. At $x=c$, we find that $f'(c) = 0$ and $f''(c) = 0$. Which of the following conclusions is logically valid?
   Answer: The Second Derivative Test is inconclusive regarding local extrema at $x=c$. When $f'(c) = 0$ and $f''(c) = 0$, the Second Derivative Test fails to provide information about whether $f$ has a local extremum at $c$. For example, if $f(x)=x^4$, then $f'(0)=0$ and $f''(0)=0$, and $f$ has a local minimum at $x=0$. If $g(x)=x^3$, then $g'(0)=0$ and $g''(0)=0$, and $g$ has no extremum at $x=0$. Because different outcomes are possible, the only valid conclusion is that the test is inconclusive.
8. Let $f(x)$ be a piecewise function defined as $f(x) = \begin{cases} x^2+4x+3 & \text{if } x \leq -1 \\ -x-1 & \text{if } x > -1 \end{cases}$. What is the sum of the x-coordinates of all local extrema of $f$?
   Answer: $-3$ To find local extrema, we must find all critical points of $f$. We check where $f'(x)=0$ and where $f'(x)$ is undefined. 1. Differentiate each piece: For $x < -1$, $f'(x) = 2x+4$. Setting this to zero gives $2x+4=0 \implies x=-2$. Since $-2 < -1$, this is a valid critical point. For $x > -1$, $f'(x) = -1$. This is never zero, so there are no critical points in this region. 2. Check the point where the definition changes, $x=-1$. A critical point exists here if the derivative is undefined. First, check for continuity: $\lim_{x \to -1^-} (x^2+4x+3) = (-1)^2+4(-1)+3 = 0$. $\lim_{x \to -1^+} (-x-1) = -(-1)-1 = 0$. Since the limits are equal, the function is continuous. Now check the one-sided derivatives: Left-hand derivative: $f'(-1^-) = 2(-1)+4 = 2$. Right-hand derivative: $f'(-1^+) = -1$. Since the left-hand and right-hand derivatives are not equal, $f'(-1)$ is undefined. Thus, $x=-1$ is a critical point. 3. The critical points are $x=-2$ and $x=-1$. We classify them using the first derivative test: - For $x<-2$, $f'(x)$ is negative (e.g., $f'(-3)=-2$). For $-2<x<-1$, $f'(x)$ is positive (e.g., $f'(-1.5)=1$). Thus, $f$ has a local minimum at $x=-2$. - For $-2<x<-1$, $f'(x)$ is positive. For $x>-1$, $f'(x)$ is negative (it's -1). Thus, $f$ has a local maximum at $x=-1$. Both critical points correspond to local extrema. The sum of their x-coordinates is $(-2) + (-1) = -3$.
9. A particle moves along the x-axis with position given by $s(t) = \frac{1}{4}t^4 - t^3 - 2t^2 + 1$ for the time interval $0 \leq t \leq 5$. At what time $t$ is the particle furthest from the origin in the negative direction?
   Answer: $t=4$ The phrase "furthest from the origin in the negative direction" means we are looking for the absolute minimum value of the position function $s(t)$ on the interval $[0, 5]$. First, find the derivative of $s(t)$, which represents the velocity $v(t)$: $v(t) = s'(t) = t^3 - 3t^2 - 4t$. Next, find the critical points by setting the derivative to zero: $t^3 - 3t^2 - 4t = 0$ $t(t^2 - 3t - 4) = 0$ $t(t-4)(t+1) = 0$ The critical times are $t=0, t=4$, and $t=-1$. We only consider the critical points within the given interval $[0, 5]$, which are $t=0$ and $t=4$. To find the absolute minimum, we test these points and the endpoint $t=5$ in the position function $s(t)$: $s(0) = \frac{1}{4}(0)^4 - (0)^3 - 2(0)^2 + 1 = 1$ $s(4) = \frac{1}{4}(4)^4 - (4)^3 - 2(4)^2 + 1 = 64 - 64 - 32 + 1 = -31$ $s(5) = \frac{1}{4}(5)^4 - (5)^3 - 2(5)^2 + 1 = \frac{625}{4} - 125 - 50 + 1 = 156.25 - 174 = -17.75$ Comparing the values $s(0)=1$, $s(4)=-31$, and $s(5)=-17.75$, the most negative position (the absolute minimum) is $-31$, which occurs at $t=4$.
10. Let $f(x) = (x^2 - 3)e^{-x}$. What is the absolute maximum value of $f(x)$ on the interval $[-2, \infty)$?
    Answer: $e^2$ The derivative is $f'(x) = (2x)e^{-x} - (x^2-3)e^{-x} = -e^{-x}(x^2-2x-3) = -e^{-x}(x-3)(x+1)$. Critical points are $x=3$ and $x=-1$. By the first derivative test, $x=-1$ is a local minimum and $x=3$ is a local maximum. The candidates for absolute maximum are the value at the local maximum and the value at the left endpoint $x=-2$. We also check the limit as $x \to \infty$. We have $f(3) = (3^2-3)e^{-3} = 6e^{-3} \approx 0.3$. At the endpoint, $f(-2) = ((-2)^2-3)e^{2} = e^2 \approx 7.39$. As $x \to \infty$, $f(x) \to 0$. Comparing these values, the absolute maximum is $e^2$.
11. Let $f(x) = |x-1|(x+2)$. Which of the following statements is true?
    Answer: $f$ has a local minimum at $x=1$ and a local maximum at $x=-1/2$. We can write $f(x)$ as a piecewise function: $f(x) = -(x-1)(x+2) = -x^2-x+2$ for $x<1$, and $f(x) = (x-1)(x+2) = x^2+x-2$ for $x \geq 1$. The derivative is $f'(x)=-2x-1$ for $x<1$ and $f'(x)=2x+1$ for $x>1$. The derivative is undefined at $x=1$, so it's a critical point. Setting $-2x-1=0$ gives $x=-1/2$, which is in $(-\infty, 1)$, so it is also a critical point. At $x=-1/2$, the derivative changes from positive (for $x<-1/2$) to negative, indicating a local maximum. At $x=1$, the derivative changes from negative (for $-1/2<x<1$) to positive, indicating a local minimum.
12. Let $f$ be a function that is continuous on $[a, b]$. Which of the following is a direct and certain conclusion from the Extreme Value Theorem?
    Answer: $f$ attains both an absolute maximum value and an absolute minimum value on $[a, b]$. The Extreme Value Theorem states that if a function is continuous on a closed, bounded interval, then it must achieve an absolute maximum and an absolute minimum value on that interval. The other statements are not guaranteed. The extrema can occur at endpoints or where the derivative is undefined, not just where it's zero (A). A monotonic function like $f(x)=x$ on $[0,1]$ has no critical points in the interior (B). Statement (D) is related to Rolle's Theorem, which has additional hypotheses and a different conclusion.
13. Let $f$ be a function that is defined for all real numbers. If $f'(2)$ is undefined, which of the following statements must be true?
    Answer: $f$ must have a critical point at $x=2$. The definition of a critical point of a function $f$ is a point $x=c$ in the domain of $f$ where either $f'(c)=0$ or $f'(c)$ is undefined. The stem states that $f$ is defined for all real numbers (so $x=2$ is in the domain) and that $f'(2)$ is undefined. Therefore, by definition, $f$ must have a critical point at $x=2$. (A) is false. A critical point is a necessary condition for a local extremum, but not a sufficient one. For example, if $f(x) = (x-2)^{1/3}$, then $f'(2)$ is undefined, but $f$ has no extremum at $x=2$. (C) is false. A function can be continuous at a point where its derivative is undefined. For example, $f(x) = |x-2|$ is continuous at $x=2$, but $f'(2)$ is undefined. (D) is false. An undefined derivative can correspond to a corner or a cusp, not just a vertical tangent. For example, $f(x)=|x-2|$ has a corner at $x=2$.
14. The function $f(x) = x^3 + ax^2 + bx$ has a local minimum at $x=4$ and a local maximum at $x=0$. What is the value of $b$?
    Answer: 0 When you encounter problems about local maxima and minima, you need to use the fundamental relationship between critical points and derivatives. A function has local extrema where its derivative equals zero, and the nature of these extrema depends on the second derivative. Given that $f(x) = x^3 + ax^2 + bx$, let's find the derivative: $f'(x) = 3x^2 + 2ax + b$. Since there's a local maximum at $x = 0$ and a local minimum at $x = 4$, both points must be critical points where $f'(x) = 0$. At $x = 0$: $f'(0) = 3(0)^2 + 2a(0) + b = b = 0$ At $x = 4$: $f'(4) = 3(16) + 2a(4) + b = 48 + 8a + b = 0$ Since $b = 0$ from the first equation, substituting into the second gives us $48 + 8a = 0$, so $a = -6$. We can verify this works by checking the second derivative: $f''(x) = 6x + 2a = 6x - 12$. At $x = 0$, $f''(0) = -12 < 0$ (confirming a maximum), and at $x = 4$, $f''(4) = 12 > 0$ (confirming a minimum). Answer choice A) -24 likely comes from miscalculating the derivative conditions. Choice B) is incorrect because we have enough information from two critical points to solve for both unknowns. Choice C) 6 might result from sign errors in the algebra. Choice D) 0 is correct because the critical point condition at $x = 0$ directly gives us $b = 0$. Study tip: Always use both given critical points to set up a system of equations—each critical point provides one equation where $f'(x) = 0$.
15. Let $f$ be a function such that $f'(c) = 0$. Which of the following additional conditions is sufficient to conclude that $f$ has a local minimum at $x=c$?
    Answer: $f'(x)$ changes from negative to positive at $x=c$. This question asks for a sufficient condition, which is a key part of a calculus test for extrema. The First Derivative Test states that if $f'(c)=0$ and $f'(x)$ changes sign from negative on the left of $c$ to positive on the right of $c$, then $f$ has a local minimum at $c$. This describes the function changing from decreasing to increasing. Choice D is the definition of a local minimum, not a test condition. Choice A is insufficient because if $f''(c)=0$, the test is inconclusive. Choice B describes a condition for an inflection point, not an extremum.
16. Let $f(x) = (x-2)^{2/3}(2x+1)$. What is the sum of the x-coordinates of all critical points of $f$?
    Answer: $3$ To find the critical points of $f(x)$, we need to find its derivative, $f'(x)$, and identify where $f'(x)=0$ or $f'(x)$ is undefined. The domain of $f(x)$ is all real numbers. Using the product rule and chain rule: $f'(x) = \frac{d}{dx}[(x-2)^{2/3}] \cdot (2x+1) + (x-2)^{2/3} \cdot \frac{d}{dx}[2x+1]$ $f'(x) = \frac{2}{3}(x-2)^{-1/3}(1) \cdot (2x+1) + (x-2)^{2/3} \cdot (2)$ To simplify, find a common denominator: $f'(x) = \frac{2(2x+1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3} = \frac{2(2x+1) + 2(x-2)^{2/3} \cdot 3(x-2)^{1/3}}{3(x-2)^{1/3}}$ $f'(x) = \frac{4x+2 + 6(x-2)}{3(x-2)^{1/3}} = \frac{4x+2+6x-12}{3(x-2)^{1/3}} = \frac{10x-10}{3(x-2)^{1/3}}$ Critical points occur where the numerator is zero or the denominator is zero. 1. $f'(x)=0$: $10x-10=0 \implies x=1$. 2. $f'(x)$ is undefined: $3(x-2)^{1/3}=0 \implies x=2$. Both $x=1$ and $x=2$ are in the domain of $f(x)$, so both are critical points. The sum of their x-coordinates is $1+2=3$.
17. Let $f(x) = x\sqrt{4-x^2}$ on its domain $[-2, 2]$. What is the difference between the absolute maximum and absolute minimum values of $f$?
    Answer: 4 First, find the derivative: $f'(x) = \sqrt{4-x^2} - \frac{x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}}$. Critical points occur when $4-2x^2=0$, which gives $x = \pm\sqrt{2}$. We must test these points and the endpoints $x=\pm 2$. $f(-2)=0$, $f(-\sqrt{2}) = -2$, $f(\sqrt{2}) = 2$, $f(2)=0$. The absolute maximum value is 2, and the absolute minimum value is -2. The difference is $2 - (-2) = 4$.
18. How many critical points does the function $f(x) = (x^2 - 4)^{\frac{2}{3}}$ have?
    Answer: 3 To find critical points, we compute the derivative: $f'(x) = \frac{2}{3}(x^2-4)^{-\frac{1}{3}}(2x) = \frac{4x}{3\sqrt[3]{x^2-4}}$. Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. $f'(x) = 0$ when the numerator is zero, so $4x=0 \implies x=0$. $f'(x)$ is undefined when the denominator is zero, so $x^2-4=0 \implies x=2$ and $x=-2$. Since the original function $f(x)$ is defined at all three of these points, there are a total of 3 critical points.
19. Let $f$ be a continuous function on the closed interval $[0, 5]$. The only critical point of $f$ in the open interval $(0, 5)$ is at $x=2$. If $f(0)=1, f(2)=4,$ and $f(5)=3$, what can be concluded about the absolute extrema of $f$ on $[0, 5]$?
    Answer: The absolute minimum is 1 and the absolute maximum is 4. According to the Extreme Value Theorem, a continuous function on a closed interval must have an absolute maximum and an absolute minimum. These extrema can only occur at critical points within the interval or at the endpoints of the interval. The possible locations for the absolute extrema are the endpoints $x=0$ and $x=5$, and the critical point $x=2$. We are given the function values at these three candidate points: $f(0) = 1$ $f(2) = 4$ $f(5) = 3$ To find the absolute maximum and minimum on the interval $[0, 5]$, we simply compare these values. The largest value is 4, which occurs at $x=2$. So, the absolute maximum is 4. The smallest value is 1, which occurs at $x=0$. So, the absolute minimum is 1. Because the problem states that $x=2$ is the only critical point in the open interval, we have a complete list of all possible locations for the absolute extrema. Therefore, we can definitively determine both.
20. The derivative of a function $f$ is given by $f'(x) = (x-3)^2(x+1)(x^2+4)$. At which value(s) of $x$ does $f$ have a local minimum?
    Answer: $x = -1$ only Local minima occur at critical points where the derivative $f'(x)$ changes sign from negative to positive. The critical points are the roots of $f'(x)=0$, which are $x=3$ and $x=-1$. The factors $(x-3)^2$ and $(x^2+4)$ are always non-negative, so they do not affect the sign change. The sign of $f'(x)$ is determined by the factor $(x+1)$. For $x < -1$, $x+1$ is negative, so $f'(x) < 0$. For $x > -1$, $x+1$ is positive, so $f'(x) > 0$. Since the sign changes from negative to positive at $x=-1$, $f$ has a local minimum there. At $x=3$, there is no sign change, so it is not an extremum.

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