First Derivative Test - Calculus 1

The function f is defined on the closed interval [-2, 2] and its derivative is given by f'(x) = $e^{x^2}$ ( $x^2$ - 3x). Find the x-coordinate of the local maximum of f in the interior of the interval.
Answer: x = 0 To find local extrema in the interior, we find critical points by setting f'(x) = 0. Since $e^{x^2}$ > 0 always, we solve $x^2$ - 3x = 0, giving x(x-3) = 0, so x = 0 or x = 3. Only x = 0 lies in the interior of [-2, 2]. Using the First Derivative Test: For x ∈ (-2, 0), both x < 0 and (x-3) < 0, so f'(x) = $e^{x^2}$ · x(x-3) > 0. For x ∈ (0, 2), x > 0 and (x-3) < 0, so f'(x) = $e^{x^2}$ · x(x-3) < 0. Since f'(x) changes from positive to negative at x = 0, there is a local maximum at x = 0.
The derivative of a function $g$ is given by $g'(x) = (x^3-4x)\ln(x^2+1)$ . At which of the following x-values does $g$ have a relative maximum?
Answer: $x=0$ only When you're asked to find relative maxima given a derivative, you need to locate critical points where the derivative equals zero, then use the second derivative test or analyze sign changes to determine whether each critical point is a maximum, minimum, or neither. First, find where $g'(x) = (x^3-4x)\ln(x^2+1) = 0$ . Since $\ln(x^2+1)$ is always positive (because $x^2+1 \geq 1$ for all real $x$ ), we only need $(x^3-4x) = 0$ . Factoring gives $x(x^2-4) = x(x-2)(x+2) = 0$ , so the critical points are $x = -2, 0, 2$ . To determine which are maxima, analyze how $g'(x)$ changes sign around each critical point. The factor $\ln(x^2+1)$ is always positive, so the sign depends entirely on $(x^3-4x)$ : - For $x < -2$ : $(x^3-4x) < 0$ , so $g'(x) < 0$ - For $-2 < x < 0$ : $(x^3-4x) > 0$ , so $g'(x) > 0$ - For $0 < x < 2$ : $(x^3-4x) < 0$ , so $g'(x) < 0$ - For $x > 2$ : $(x^3-4x) > 0$ , so $g'(x) > 0$ A relative maximum occurs when the derivative changes from positive to negative. This happens only at $x = 0$ (positive to negative). At $x = -2$ and $x = 2$ , the derivative changes from negative to positive, indicating relative minima. Choice A incorrectly identifies the minima as maxima. Choice B misses that $x = 2$ is a minimum, not a maximum. Choice D incorrectly includes all critical points without checking the sign changes. Remember: always verify whether critical points are maxima or minima by checking how the derivative's sign changes around each point.
Let $f(x) = x^4 - 4x^3 + 4x^2 + 1$ . Which of the following statements correctly classifies all the relative extrema of $f$ ?
Answer: Relative minima at $x=0$ and $x=2$ ; a relative maximum at $x=1$ . When you encounter a question about classifying relative extrema, you need to find critical points using the first derivative, then determine whether each is a maximum or minimum using the second derivative test. Start by finding $f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x-1)(x-2)$ . Setting this equal to zero gives critical points at $x = 0, 1, 2$ . Next, use the second derivative: $f''(x) = 12x^2 - 24x + 8$ . Now test each critical point: - At $x = 0$ : $f''(0) = 8 > 0$ , so this is a relative minimum - At $x = 1$ : $f''(1) = 12 - 24 + 8 = -4 < 0$ , so this is a relative maximum - At $x = 2$ : $f''(2) = 48 - 48 + 8 = 8 > 0$ , so this is a relative minimum This confirms answer B: relative minima at $x = 0$ and $x = 2$ , with a relative maximum at $x = 1$ . Answer A incorrectly swaps the classifications—it claims maxima where there are actually minima and vice versa. This often happens when students confuse the second derivative test rule. Answer C ignores the minima at $x = 0$ and $x = 2$ , while answer D completely reverses the situation by claiming no maxima exist when there's clearly one at $x = 1$ . Remember: when $f''(x) > 0$ at a critical point, you have a relative minimum (the graph is concave up, like a bowl). When $f''(x) < 0$ , you have a relative maximum (concave down, like an upside-down bowl).
Let $f(x) = x^2 e^{-ax}$ where $a$ is a positive constant. The function $f$ has a relative extremum for some $x>0$ . What is the location and type of this extremum?
Answer: A relative maximum at $x = 2/a$ . To find the relative extrema, we first find the derivative of $f(x)$ using the product rule: $f'(x) = (2x)(e^{-ax}) + (x^2)(-ae^{-ax})$ . Factor out common terms: $f'(x) = xe^{-ax}(2-ax)$ . Next, find the critical points by setting $f'(x)=0$ . Since $e^{-ax}$ is always positive, we only need to solve $x(2-ax)=0$ . This gives $x=0$ and $2-ax=0$ , which means $x=2/a$ . The problem specifies $x>0$ , so we are interested in the critical point $x=2/a$ . (Since $a>0$ , $2/a$ is positive). Now, we apply the First Derivative Test to the point $x=2/a$ . We analyze the sign of $f'(x) = xe^{-ax}(2-ax)$ in an interval around $2/a$ . For $x>0$ , both $x$ and $e^{-ax}$ are positive, so the sign of $f'(x)$ is determined by the sign of the term $(2-ax)$ . - When $0 < x < 2/a$ , we have $ax < 2$ , so $2-ax > 0$ . Thus, $f'(x)$ is positive. - When $x > 2/a$ , we have $ax > 2$ , so $2-ax < 0$ . Thus, $f'(x)$ is negative. Since $f'(x)$ changes from positive to negative at $x=2/a$ , the function $f(x)$ has a relative maximum at this point.
Let f be a differentiable function with f'(x) = $x^2$ - 4. Let g(x) = f(2x-3). At which x-value does g(x) have a local minimum?
Answer: x = 5/2 We use the chain rule to find g'(x): g'(x) = f'(2x-3) \cdot 2. The critical points of g(x) are where g'(x) = 0. This means f'(2x-3) = 0. From the definition of f', we have f'(u) = $u^2$ - 4. So we need to solve $(2x-3)^2$ - 4 = 0. This gives $(2x-3)^2$ = 4, so 2x-3 = 2 or 2x-3 = -2. Solving for x, we get 2x = 5 \implies x = 5/2, and 2x = 1 \implies x = 1/2. Now we test these critical points using the First Derivative Test for g(x). The sign of g'(x) is determined by the sign of f'(2x-3) = $(2x-3)^2$ - 4. This is an upward-opening parabola in the variable (2x-3), and also an upward-opening parabola in x. The roots are x=1/2 and x=5/2. Thus, g'(x) is positive for x < 1/2, negative for 1/2 < x < 5/2, and positive for x > 5/2. At x=1/2, g'(x) changes from positive to negative, which is a local maximum. At x=5/2, g'(x) changes from negative to positive, which is a local minimum.
Let $f(x) = x(x-4)^3$ . If $g(x) = |f(x)|$ , at which x-values does $g(x)$ have relative minima?
Answer: At $x=0$ and $x=4$ only When you see absolute value functions and need to find relative minima, you're looking for points where the function changes from decreasing to increasing. The key insight is that $g(x) = |f(x)|$ will have relative minima wherever $f(x) = 0$ (creating sharp corners) or wherever $f(x)$ itself has relative minima that remain below the x-axis. First, let's analyze $f(x) = x(x-4)^3$ . To find critical points, we need $f'(x) = (x-4)^3 + x \cdot 3(x-4)^2 = (x-4)^2[(x-4) + 3x] = (x-4)^2(4x-4) = 4(x-4)^2(x-1)$ . Setting $f'(x) = 0$ gives us $x = 1$ and $x = 4$ . Next, find where $f(x) = 0$ : $x(x-4)^3 = 0$ gives us $x = 0$ and $x = 4$ . These are crucial because $|f(x)|$ creates sharp points here. Now analyze the behavior: $f(x) < 0$ when $0 < x < 4$ , and $f(x) > 0$ when $x < 0$ or $x > 4$ . At $x = 0$ and $x = 4$ , the function $g(x) = |f(x)|$ touches zero and changes direction, creating relative minima. At $x = 1$ , even though $f(x)$ has a relative maximum, since $f(1) = 1(-3)^3 = -27 < 0$ , we have $g(1) = 27$ , which is actually a relative maximum for $g(x)$ , not a minimum. Choice A incorrectly includes $x = 1$ . Choice B only identifies $x = 1$ , missing the actual minima. Choice D includes all critical points without considering the behavior of the absolute value. Strategy tip: For absolute value problems, always identify where the inside function equals zero and check whether critical points create maxima or minima after applying absolute value.
Let f(x) be a continuous function. If f'(x) > 0 for x < 2 and f'(x) < 0 for x > 2, but f'(2) is undefined, what must be true about f at x=2?
Answer: f has a local maximum at x=2. The First Derivative Test applies to critical points where the derivative is zero or undefined. Here, x=2 is a critical point because f'(2) is undefined. The test examines the sign of the derivative on either side of the critical point. We are given that f'(x) is positive for x < 2, which means f is increasing. We are given that f'(x) is negative for x > 2, which means f is decreasing. Since the function changes from increasing to decreasing at x=2, f must have a local maximum at x=2. This could be a cusp or a corner point.
The derivative of a function $f$ is given by $f'(x) = (x-a)^2 (x-b)$ for constants $a$ and $b$ where $a \neq b$ . Which of the following statements must be true about the relative extrema of $f$ ?
Answer: $f$ has a relative minimum at $x=b$ . When you encounter a derivative and need to analyze extrema, you're looking for critical points where the derivative equals zero, then determining whether each critical point is a maximum, minimum, or neither using the first or second derivative test. Given $f'(x) = (x-a)^2(x-b)$ , the critical points occur when $f'(x) = 0$ . This happens when $(x-a)^2 = 0$ or $(x-b) = 0$ , giving us critical points at $x = a$ and $x = b$ . To determine the nature of each critical point, let's analyze the sign of $f'(x)$ around these points. Since $(x-a)^2$ is always non-negative (it's a perfect square), the sign of $f'(x)$ depends entirely on the sign of $(x-b)$ . When $x < b$ , we have $(x-b) < 0$ , so $f'(x) \leq 0$ . When $x > b$ , we have $(x-b) > 0$ , so $f'(x) \geq 0$ . This means $f'(x)$ changes from negative to positive at $x = b$ , indicating a relative minimum there. Now examining each choice: (A) is incorrect because at $x = a$ , the factor $(x-a)^2$ equals zero, but since it's squared, $f'(x)$ doesn't change sign around $x = a$ – it remains the same sign as $(x-b)$ on both sides. (B) is wrong because we showed $x = b$ gives a minimum, not maximum. (D) is incorrect because the relative minimum at $x = b$ occurs regardless of whether $a > b$ or $b > a$ . Study tip: When analyzing critical points, always check whether the derivative actually changes sign – squared factors don't cause sign changes, so they don't create extrema.
The derivative of a function $f$ is given by $f'(x) = (x-k)(x+2)$ . The function $f$ is known to have a relative minimum at $x=3$ . What is the nature of the function's other critical point at $x=-2$ ?
Answer: A relative maximum For $f$ to have a relative minimum at $x=3$ , its derivative $f'(x)$ must be zero at $x=3$ and change sign from negative to positive. The critical points of $f'(x) = (x-k)(x+2)$ are $x=k$ and $x=-2$ . Since one of the critical points is at $x=3$ , we must have $k=3$ . Thus, $f'(x) = (x-3)(x+2)$ . The critical points are $x=3$ and $x=-2$ . Let's test the sign of $f'(x)$ around $x=-2$ . For $x<-2$ (e.g., $x=-3$ ), $f'(-3) = (-6)(-1) > 0$ . For $-2<x<3$ (e.g., $x=0$ ), $f'(0) = (-3)(2) < 0$ . Since $f'(x)$ changes from positive to negative at $x=-2$ , the function $f$ has a relative maximum at $x=-2$ .
Let f'(x) = $(x-a)^3$ $(x-b)^2$ for constants a < b. Which of the following correctly describes the local extrema of f(x)?
Answer: Local minimum at x=a, no extremum at x=b. When you're given a derivative and asked about extrema of the original function, you need to analyze where the derivative equals zero and determine the nature of each critical point using the first derivative test. Given $f'(x) = (x-a)^3(x-b)^2$ , the critical points occur where $f'(x) = 0$ , which happens at $x = a$ and $x = b$ . To classify each critical point, examine how the sign of $f'(x)$ changes around each point. At $x = a$ : The factor $(x-a)^3$ has an odd exponent, so it changes sign as you cross $x = a$ . Since $(x-b)^2$ is always positive (even exponent) and $a < b$ , we have $f'(x) < 0$ for $x < a$ and $f'(x) > 0$ for $x > a$ . This sign change from negative to positive indicates a local minimum. At $x = b$ : The factor $(x-b)^2$ has an even exponent, so it doesn't change sign across $x = b$ – it's positive on both sides. Since $(x-a)^3 > 0$ for $x > a$ , and $b > a$ , we have $f'(x) > 0$ both before and after $x = b$ . No sign change means no extremum. Option A incorrectly identifies $x = a$ as a maximum and claims an extremum at $x = b$ . Option B makes the same error about $x = b$ while also misidentifying the minimum. Option C correctly identifies the minimum at $x = a$ but wrongly claims a maximum at $x = b$ . Key insight: The exponent determines behavior – odd exponents create sign changes (extrema), while even exponents don't change signs (no extrema).
Let $f(x) = \frac{\ln x}{x^2}$ for $x>0$ . At which value of $x$ does $f(x)$ have a relative extremum?
Answer: A relative maximum at $x = \sqrt{e}$ First, find the derivative using the quotient rule: $f'(x) = \frac{(\frac{1}{x})x^2 - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2 \ln x}{x^3}$ . To find critical points, set $f'(x) = 0$ . Since $x>0$ , we only need to solve $1 - 2 \ln x = 0$ , which gives $2 \ln x = 1$ , or $\ln x = 1/2$ , so $x = e^{1/2} = \sqrt{e}$ . To classify this extremum, we use the First Derivative Test. For $0 < x < \sqrt{e}$ , choose $x=1$ . $f'(1) = \frac{1-2\ln 1}{1^3} = 1 > 0$ . For $x > \sqrt{e}$ , choose $x=e$ . $f'(e) = \frac{1-2\ln e}{e^3} = \frac{1-2}{e^3} < 0$ . Since the derivative changes from positive to negative, $f$ has a relative maximum at $x = \sqrt{e}$ .
A function f is twice-differentiable. It is known that f'(c)=0. To determine if f has a local extremum at x=c using the First Derivative Test, which of the following is sufficient?
Answer: f'(x) changes sign at x=c. The First Derivative Test for a local extremum at a critical point c states that if the derivative f'(x) changes sign at x=c, then f has a local extremum at c. Choice A is the condition for the Second Derivative Test, not the First. Choice C describes the conditions for a local minimum specifically, which is sufficient for an extremum, but it is not the most general condition for an extremum (which could be a maximum or a minimum). Choice B is the core condition of the test for the existence of any local extremum. Choice D is a necessary condition for differentiability at c (which implies f'(c) exists), but it is not sufficient to guarantee an extremum.
Let $f(x) = x - 2\cos(x)$ on the interval $[0, 2\pi]$ . At which value of $x$ does $f(x)$ have a relative minimum?
Answer: $x=\frac{11\pi}{6}$ To find relative extrema, we first compute the derivative of $f(x)$ : $f'(x) = 1 - 2(-\sin(x)) = 1 + 2\sin(x)$ . Next, we find the critical points by setting $f'(x)=0$ for $x$ in $[0, 2\pi]$ . $1 + 2\sin(x) = 0$ $2\sin(x) = -1$ $\sin(x) = -\frac{1}{2}$ The solutions for $x$ in the interval $[0, 2\pi]$ are $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$ . Now we use the First Derivative Test to classify these critical points. We check the sign of $f'(x) = 1+2\sin(x)$ on the intervals created by these points: $[0, \frac{7\pi}{6})$ , $(\frac{7\pi}{6}, \frac{11\pi}{6})$ , and $(\frac{11\pi}{6}, 2\pi]$ . - Interval $[0, \frac{7\pi}{6})$ : Let's test $x=\pi$ . $f'(\pi) = 1 + 2\sin(\pi) = 1+0=1 > 0$ . - Interval $(\frac{7\pi}{6}, \frac{11\pi}{6})$ : Let's test $x=\frac{3\pi}{2}$ . $f'(\frac{3\pi}{2}) = 1 + 2\sin(\frac{3\pi}{2}) = 1+2(-1) = -1 < 0$ . - Interval $(\frac{11\pi}{6}, 2\pi]$ : Let's test $x=2\pi$ . $f'(2\pi) = 1 + 2\sin(2\pi) = 1+0=1 > 0$ . At $x=\frac{7\pi}{6}$ , $f'(x)$ changes from positive to negative, indicating a relative maximum. At $x=\frac{11\pi}{6}$ , $f'(x)$ changes from negative to positive, indicating a relative minimum. The question asks for the relative minimum, which occurs at $x=\frac{11\pi}{6}$ .
Let $f$ be a function with derivative $f'(x) = (x^2 - kx)(x-4)$ , where $k$ is a constant. If $f$ has a relative maximum at $x=2$ , which of the following statements about $f$ is true?
Answer: $f$ has a relative minimum at $x=0$ . For $f$ to have a relative extremum at $x=2$ , $x=2$ must be a critical point, so $f'(2)=0$ . Substitute $x=2$ into the expression for $f'(x)$ : $f'(2) = (2^2 - k(2))(2-4) = (4-2k)(-2) = 0$ . Solving for $k$ gives $4-2k=0$ , so $k=2$ . Now, substitute $k=2$ back into the derivative: $f'(x) = (x^2-2x)(x-4) = x(x-2)(x-4)$ . The critical points are the roots of $f'(x)$ , which are $x=0$ , $x=2$ , and $x=4$ . To determine the nature of the extrema, we use the First Derivative Test. We test the sign of $f'(x)$ in the intervals defined by the critical points: - For $x<0$ (e.g., $x=-1$ ), $f'(-1) = (-1)(-3)(-5) < 0$ . $f$ is decreasing. - For $0<x<2$ (e.g., $x=1$ ), $f'(1) = (1)(-1)(-3) > 0$ . $f$ is increasing. - For $2<x<4$ (e.g., $x=3$ ), $f'(3) = (3)(1)(-1) < 0$ . $f$ is decreasing. - For $x>4$ (e.g., $x=5$ ), $f'(5) = (5)(3)(1) > 0$ . $f$ is increasing. At $x=0$ , the derivative changes from negative to positive, so $f$ has a relative minimum at $x=0$ . At $x=2$ , the derivative changes from positive to negative, confirming the given relative maximum. At $x=4$ , the derivative changes from negative to positive, so $f$ has a relative minimum at $x=4$ . Therefore, the statement that $f$ has a relative minimum at $x=0$ is true.
For the function f(x) = $x^4$ - $4x^3$ , which statement is correct?
Answer: f has a local minimum at x=3 and no local maximum. The derivative is f'(x) = $4x^3$ - $12x^2$ = $4x^2$ (x-3). The critical points are where f'(x)=0, which gives x=0 and x=3. We apply the First Derivative Test. For x<0, $4x^2$ is positive and (x-3) is negative, so f'(x) is negative. For 0<x<3, 4x^2 is positive and (x-3) is negative, so f'(x) is negative. At x=0, the sign of f'(x) does not change, so it is neither a local maximum nor a local minimum. For x>3, $4x^2$ is positive and (x-3) is positive, so f'(x) is positive. At x=3, the sign of f'(x) changes from negative to positive, indicating a local minimum. Therefore, the function has a local minimum at x=3 and no local maximum.
A function f is defined by f(x) = $x^3$ - $6x^2$ + ax + b, where a and b are constants. If f has a local minimum at x = 3, what is the value of its local maximum?
Answer: f(1) The derivative is f'(x) = $3x^2$ - 12x + a. Since f has a local minimum at x = 3, x=3 must be a critical point, so f'(3) = 0. Plugging in x=3: 3 $(3)^2$ - 12(3) + a = 0 \implies 27 - 36 + a = 0 \implies a = 9. So, f'(x) = $3x^2$ - 12x + 9 = 3( $x^2$ - 4x + 3) = 3(x-1)(x-3). The critical points are x=1 and x=3. The problem states x=3 is a local minimum. Let's check the sign of f'(x). It's an upward-opening parabola with roots at 1 and 3. So f' is positive for x<1, negative for 1<x<3, and positive for x>3. At x=1, f' changes from positive to negative, indicating a local maximum. At x=3, f' changes from negative to positive, indicating a local minimum, which is consistent with the problem statement. The local maximum occurs at x=1, and its value is f(1).
Let $f$ be a function such that its derivative is $f'(x) = (x-a)^n g(x)$ , where $n$ is a positive integer, $a$ is a constant, and $g$ is a continuous function with $g(a) > 0$ . Under which condition will $f$ have a relative minimum at $x=a$ ?
Answer: If $n$ is any odd positive integer When you encounter a problem about relative extrema using the first derivative, you need to analyze how the sign of $f'(x)$ changes around the critical point where $f'(a) = 0$ . Since $f'(x) = (x-a)^n g(x)$ and $g(a) > 0$ with $g$ continuous, the function $g(x)$ remains positive in some neighborhood around $x = a$ . This means the sign of $f'(x)$ near $x = a$ depends entirely on the factor $(x-a)^n$ . For a relative minimum, you need $f'(x)$ to change from negative to positive as $x$ increases through $a$ . When $n$ is odd, $(x-a)^n$ is negative for $x < a$ and positive for $x > a$ . Since $g(x) > 0$ throughout this interval, $f'(x)$ changes from negative to positive, creating a relative minimum at $x = a$ . Choice A is incorrect because when $n$ is even, $(x-a)^n$ is positive on both sides of $x = a$ , so $f'(x)$ doesn't change sign—there's no extremum. Choice C is wrong because, as explained above, even values of $n$ don't produce relative minima. Choice D is irrelevant because the behavior depends on the power $n$ , not on properties of $g'(a)$ . The continuity of $g$ and the fact that $g(a) > 0$ are sufficient. Study tip: Remember that for critical points of the form $(x-a)^n$ , odd powers create extrema while even powers create neither maxima nor minima (they're inflection points or flat regions).
The derivative of the function f is given by f'(x) = $e^x$ ( $x^2-1$ ) - $e^x$ (2x). Where does f have a local maximum?
Answer: x = 1 - \sqrt{2} First, simplify the derivative: f'(x) = $e^x$ ( $x^2$ - 1 - 2x) = $e^x$ ( $x^2$ - 2x - 1). The critical points occur when f'(x) = 0. Since $e^x$ is never zero, we must solve $x^2$ - 2x - 1 = 0. Using the quadratic formula, x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}. The critical points are x = 1 - \sqrt{2} and x = 1 + \sqrt{2}. The sign of f'(x) is determined by the quadratic $x^2$ - 2x - 1, which is an upward-opening parabola. Thus, f'(x) is positive on ( $-\infty, 1-\sqrt{2}$ ), negative on ( $1-\sqrt{2}, 1+\sqrt{2}$ ), and positive on ( $1+\sqrt{2}, \infty$ ). At x = 1 - \sqrt{2}, the derivative changes from positive to negative, indicating a local maximum. At x = 1 + \sqrt{2}, the derivative changes from negative to positive, indicating a local minimum.
Consider the function $f(x) = \frac{x^2+3}{x-1}$ . Which of the following statements correctly describes the relative extrema of $f$ ?
Answer: $f$ has a relative maximum at $x=-1$ and a relative minimum at $x=3$ . When finding relative extrema for rational functions, you need to locate critical points by finding where the derivative equals zero or is undefined, then analyze the behavior around those points. For $f(x) = \frac{x^2+3}{x-1}$ , let's find $f'(x)$ using the quotient rule: $f'(x) = \frac{(x-1)(2x) - (x^2+3)(1)}{(x-1)^2} = \frac{2x^2-2x-x^2-3}{(x-1)^2} = \frac{x^2-2x-3}{(x-1)^2}$ Setting the numerator equal to zero: $x^2-2x-3 = 0$ , which factors as $(x-3)(x+1) = 0$ . This gives critical points at $x = -1$ and $x = 3$ . Note that $x = 1$ makes the derivative undefined, but it's also where the original function is undefined (vertical asymptote), so it's not a candidate for a relative extremum. Testing the sign of $f'(x)$ around our critical points: for $x < -1$ , $f'(x) > 0$ ; for $-1 < x < 1$ , $f'(x) < 0$ ; for $1 < x < 3$ , $f'(x) < 0$ ; for $x > 3$ , $f'(x) > 0$ . Since $f'(x)$ changes from positive to negative at $x = -1$ , this is a relative maximum. Since $f'(x)$ changes from negative to positive at $x = 3$ , this is a relative minimum. Choice A is wrong because $x = 1$ is not in the domain. Choice B incorrectly identifies $x = -1$ as a minimum. Choice C incorrectly claims there's an extremum at the undefined point $x = 1$ . Choice D correctly identifies the relative maximum at $x = -1$ and relative minimum at $x = 3$ . Remember: always check that critical points are actually in the function's domain before classifying them as extrema.
Let $g(x) = (x^2 - 4)^{2/3}$ . Which of the following correctly describes all relative extrema of the function $g$ ?
Answer: A relative maximum at $x=0$ and relative minima at $x=-2$ and $x=2$ . To find relative extrema, you need to locate critical points where the derivative equals zero or is undefined, then determine whether each point is a maximum, minimum, or neither. First, find the derivative using the chain rule: $g'(x) = \frac{2}{3}(x^2 - 4)^{-1/3} \cdot 2x = \frac{4x}{3(x^2 - 4)^{1/3}}$ Critical points occur when $g'(x) = 0$ or is undefined. The derivative equals zero when $x = 0$ , and it's undefined when $x^2 - 4 = 0$ , giving us $x = -2$ and $x = 2$ . Now analyze the behavior around each critical point. For $x = 0$ : when $x < 0$ , we have $g'(x) < 0$ (negative slope), and when $x > 0$ , we have $g'(x) > 0$ (positive slope). This indicates a relative minimum at $x = 0$ . Wait - let me recalculate this more carefully. At $x = 0$ : $g'(x) = \frac{4x}{3(-4)^{1/3}} = \frac{4x}{3(-4)^{1/3}}$ . For $x$ slightly less than 0, $g'(x) > 0$ , and for $x$ slightly greater than 0, $g'(x) < 0$ . This means we have a relative maximum at $x = 0$ . At $x = -2$ and $x = 2$ , the derivative is undefined, but the function value is $g(-2) = g(2) = 0$ , which are the lowest points on the graph, making them relative minima. Choice A ignores the critical point at $x = 0$ . Choice B incorrectly identifies the nature of all three extrema. Choice C misses the minima at $x = ±2$ . Study tip: When derivatives are undefined due to fractional exponents, don't forget to check if those points are still extrema by examining the function's behavior around them.

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