First Derivative Test - Business Calculus

A manufacturing cost function is $C(q) = q^4 - 8q^3 + 18q^2 + 5$ where $q$ is the production level. The derivative is $C'(q) = 4q(q-1)(q-3)$ . Using the first derivative test, what can be concluded about the nature of the critical points?
Answer: $q = 0$ is neither maximum nor minimum, $q = 1$ is a local minimum, and $q = 3$ is a local maximum of the cost function Testing signs around each critical point: For q=0, C'(-0.1)=4(-0.1)(-1.1)(-3.1)<0 and C'(0.1)=4(0.1)(-0.9)(-2.9)>0, but since q=0 is a boundary in the business context, it's neither a local max nor min in the usual sense. For q=1, C'(0.5)<0 and C'(1.5)>0, so it's a local minimum. For q=3, C'(2.5)>0 and C'(3.5)<0, so it's a local maximum. Choice A incorrectly classifies q=0 and q=3. Choice B incorrectly classifies all three points. Choice C incorrectly dismisses q=0 and misclassifies q=3.
The profit, $P(x)$ , in thousands of dollars, from producing $x$ thousand units of a product is given by $P(x) = \frac{1}{3}x^3 - 9x^2 + 80x - 100$ for $x \ge 0$ . The production manager identifies a critical point at $x=10$ . Using the First Derivative Test, which of the following correctly describes the company's profit at this production level?
Answer: Profit is at a local minimum because it changes from decreasing to increasing at $x=10$ . The profit function is $P(x) = \frac{1}{3}x^3 - 9x^2 + 80x - 100$ . The marginal profit function is the first derivative, $P'(x) = x^2 - 18x + 80$ . Factoring the derivative gives $P'(x) = (x-8)(x-10)$ . The critical points are $x=8$ and $x=10$ . The question asks to analyze the critical point at $x=10$ . We use the First Derivative Test. For a value slightly less than 10 (e.g., $x=9$ ), $P'(9) = (9-8)(9-10) = (1)(-1) = -1 < 0$ , so profit is decreasing. For a value slightly greater than 10 (e.g., $x=11$ ), $P'(11) = (11-8)(11-10) = (3)(1) = 3 > 0$ , so profit is increasing. Since the derivative changes from negative to positive at $x=10$ , this point corresponds to a local minimum.
The cost, $C(x)$ , of producing $x$ handcrafted items is given by $C(x) = 10x^{2/3} + 500$ for $x \ge 0$ . Which of the following statements is true regarding the critical points of the cost function?
Answer: A critical point exists at $x=0$ , where cost is at a local minimum. A critical point is a point in the domain of a function where the derivative is either zero or undefined. The cost function is $C(x) = 10x^{2/3} + 500$ . Its derivative is $C'(x) = 10 \cdot \frac{2}{3}x^{-1/3} = \frac{20}{3x^{1/3}}$ . The derivative $C'(x)$ is never equal to zero. However, the derivative is undefined when its denominator is zero, which occurs at $x=0$ . Since $x=0$ is in the domain of the original function $C(x)$ , it is a critical point. Now we apply the First Derivative Test. For $x < 0$ , this is not in the domain of production, but mathematically, $C'(x)$ would be negative. For $x > 0$ (e.g., $x=1$ ), $C'(1) = \frac{20}{3} > 0$ . This means the function is increasing for all $x>0$ . The point $x=0$ is an endpoint of the domain of interest $[0, \infty)$ and the function increases from this point, making it a local (and absolute) minimum. The value is $C(0) = 500$ .
A company's profit function is $P(x) = -2x^3 + 15x^2 - 24x + 10$ where $x$ is the number of units produced (in thousands). After finding all critical points, the first derivative test reveals that $x = 1$ is a local maximum and $x = 4$ is a local minimum. If the company can only produce between 0 and 6 thousand units due to capacity constraints, what is the maximum profit achievable?
Answer: $P(6) = 46$ thousand dollars, so the maximum profit is $46000$ dollars The first derivative test identifies local extrema, but for a closed interval we must check critical points AND endpoints. We have local max at x=1 with P(1)=-1, local min at x=4 with P(4)=-22, plus endpoints P(0)=10 and P(6)=46. The absolute maximum is P(6)=46 thousand dollars. Choice A gives the local maximum value but ignores endpoints. Choice C gives an endpoint value but not the maximum one. Choice D gives the local minimum value.
The revenue function for a product is $R(x) = xe^{-0.1x}$ where $x$ is the advertising spend in thousands of dollars. To apply the first derivative test, you find $R'(x) = e^{-0.1x}(1 - 0.1x)$ . Which statement correctly describes the critical point analysis?
Answer: The critical point $x = 10$ is a local maximum because $R'(9) = 0.9e^{-0.9} > 0$ and $R'(11) = -0.1e^{-1.1} < 0$ Critical points occur where R'(x)=0. Since $e^{-0.1x}$ > 0 for all x, we need 1-0.1x=0, giving x=10. For the first derivative test: R'(9)= $e^{-0.9}$ (1-0.9)= $0.1e^{-0.9}$ >0 and R'(11)= $e^{-1.1}$ (1-1.1)=- $0.1e^{-1.1}$ <0. The derivative changes from positive to negative, confirming a local maximum. Choice B has incorrect signs in the derivative evaluation. Choice C incorrectly concludes no critical point exists. Choice D incorrectly identifies x=1 as the critical point.
A subscription service's churn rate is $C(p) = \frac{p^2}{100-p}$ where $p$ is the monthly price in dollars. The derivative is $C'(p) = \frac{p(200-p)}{(100-p)^2}$ . To minimize churn using the first derivative test, what price should be recommended?
Answer: $p = 0$ dollars because this makes the churn rate zero and $C'(p)$ changes from negative to positive Critical points occur where C'(p)=0: p(200-p)=0 gives p=0 or p=200. However, p=200 is outside the practical domain (p<100 to avoid division by zero). At p=0: C'(-1)<0 and C'(1)>0, so the first derivative test confirms a local minimum. While p=0 gives C(0)=0 (zero churn), this isn't realistic business-wise. The analysis suggests pricing should be as low as feasible. Choice B incorrectly focuses on the domain boundary. Choice C uses an invalid critical point. Choice D incorrectly claims the methods fail.
Let $P(x)$ be a differentiable profit function for producing $x$ units. A consultant finds that $x=500$ is the only critical point for $x>0$ . They also determine that the marginal profit $P'(400)$ is positive and the marginal profit $P'(600)$ is negative. What can be concluded about the production level $x=500$ ?
Answer: $x=500$ represents a local maximum for the profit function. When analyzing critical points in business calculus, you need to determine whether they represent profit maximization or minimization by examining the behavior of the marginal profit function around that point. Since $x = 500$ is the only critical point for $x > 0$ , we know $P'(500) = 0$ . The key insight comes from examining the sign of the derivative on either side of this critical point. You're told that $P'(400) > 0$ and $P'(600) < 0$ . This means the marginal profit is positive when producing 400 units (profit is increasing as production increases toward 500) and negative when producing 600 units (profit is decreasing as production increases beyond 500). This sign change from positive to negative as we pass through $x = 500$ confirms that the profit function reaches a local maximum at this point. Answer A is incorrect because a local minimum would require the derivative to change from negative to positive, not positive to negative. Answer B is wrong because the sign behavior of the derivative around a critical point is sufficient to classify it as a maximum or minimum—you don't need the explicit formula. Answer C misunderstands the concept; diminishing returns refers to decreasing marginal product, not the location of profit extrema, and we clearly have enough information to identify this as a local extremum. Remember: when the derivative changes from positive to negative at a critical point, you have a local maximum. This is the first derivative test in action—a powerful tool for optimizing business functions.
The marginal cost for producing a particular tablet is given by $C'(x) = \frac{10(x-8)}{\sqrt{x}}$ , where $x$ is the number of tablets produced ( $x>0$ ). Over which interval is the total cost function $C(x)$ decreasing?
Answer: The cost function is decreasing for $0 < x < 8$ . The total cost function $C(x)$ is decreasing when its derivative, the marginal cost $C'(x)$ , is negative. We are given $C'(x) = \frac{10(x-8)}{\sqrt{x}}$ . The domain is $x>0$ . To find where $C'(x) < 0$ , we need to analyze its sign. The denominator, $\sqrt{x}$ , is always positive for $x > 0$ . Therefore, the sign of $C'(x)$ is determined entirely by the sign of the numerator, $10(x-8)$ . The expression $10(x-8)$ is negative when $x-8 < 0$ , which means $x < 8$ . Combining this with the domain restriction $x>0$ , we find that $C'(x)$ is negative on the interval $(0, 8)$ . Therefore, the total cost function $C(x)$ is decreasing for $0 < x < 8$ .
The productivity function for a team is $P(h) = 4h - 0.1h^2$ where $h$ is hours worked per day. A consultant finds $P'(h) = 4 - 0.2h$ and concludes that productivity is maximized at $h = 20$ hours. When asked to verify this using the first derivative test, what should the consultant discover?
Answer: The first derivative test confirms a maximum exists, but $h = 20$ exceeds reasonable daily work hours The calculus is correct: P'(h)=4-0.2h=0 gives h=20, and P'(19)=0.2>0 while P'(21)=-0.2<0, confirming a local maximum by the first derivative test. However, 20 hours per day is unrealistic for sustainable productivity (people need sleep, breaks, etc.). The mathematical maximum occurs outside the practical domain. A reasonable constraint might be h≤12, requiring evaluation at the boundary. Choice A ignores the practical issue. Choice B incorrectly suggests the test is inconclusive. Choice D incorrectly claims P'(20)≠0.
The demand function for a luxury item is $D(p) = \frac{1000}{p+2}$ where $p$ is the price in dollars. A student claims that since $D'(p) = -\frac{1000}{(p+2)^2} < 0$ for all $p > 0$ , the first derivative test cannot be applied because there are no critical points. Which response best evaluates this reasoning?
Answer: The student is correct about no critical points, but the first derivative test can still provide information about monotonicity over intervals The student correctly identifies that D'(p)<0 for all p>0, meaning there are no critical points in the relevant domain. However, the first derivative test isn't only about finding local extrema; it also confirms that the function is strictly decreasing throughout the domain. This monotonicity information is valuable for business decisions. Choice A is partially correct but misses the broader application. Choice B incorrectly suggests p=-2 is relevant to the business context. Choice D incorrectly calculates a critical point that doesn't exist.
A company's efficiency function is $E(n) = \frac{n^3 - 6n^2 + 11n - 6}{n-1}$ where $n$ is the number of employees ( $n > 1$ ). After polynomial division, this simplifies to $E(n) = n^2 - 5n + 6$ for $n ≠ 1$ . Using $E'(n) = 2n - 5$ , what does the first derivative test reveal about efficiency optimization?
Answer: Efficiency is minimized at $n = 2.5$ employees, but since this isn't a whole number, check $n = 2$ and $n = 3$ When analyzing efficiency optimization with calculus, you need to find critical points using the first derivative, then determine whether each represents a maximum or minimum using the first derivative test. The derivative $E'(n) = 2n - 5$ equals zero when $n = 2.5$ , giving us our critical point. To apply the first derivative test, check the sign of $E'(n)$ on either side of this point. For $n < 2.5$ (say $n = 2$ ): $E'(2) = 2(2) - 5 = -1 < 0$ , so efficiency is decreasing. For $n > 2.5$ (say $n = 3$ ): $E'(3) = 2(3) - 5 = 1 > 0$ , so efficiency is increasing. Since the derivative changes from negative to positive at $n = 2.5$ , this point represents a minimum. However, since you can't hire 2.5 employees, you must evaluate the function at the nearest whole numbers to find the practical optimum. Choice A incorrectly states the direction of change - efficiency decreases before $n = 2.5$ and increases after. Choice B makes the same directional error and incorrectly suggests hiring 2.5 employees, which is impossible. Choice C wrongly claims the critical point creates a discontinuity - the simplified function $E(n) = n^2 - 5n + 6$ is continuous for all $n > 1$ , and the critical point $n = 2.5$ is well within the valid domain. Study tip: When optimization problems yield non-integer solutions in contexts requiring whole numbers (employees, products, etc.), always check the integer values on either side of the critical point to find the practical optimum.
The total profit $P(x)$ from manufacturing $x$ thousand widgets is given by $P(x) = 2x^3 - 39x^2 + 180x + 1000$ for the production range $0 \le x \le 12$ . Identify all production levels within the open interval $(0, 12)$ that correspond to local maxima of the profit function.
Answer: A local maximum occurs only at $x=3$ . To find local extrema, we first find the critical points by taking the derivative of the profit function and setting it to zero. The marginal profit is $P'(x) = 6x^2 - 78x + 180$ . Set $P'(x)=0$ : $6(x^2 - 13x + 30) = 0$ , which simplifies to $6(x-3)(x-10) = 0$ . The critical points are $x=3$ and $x=10$ , both of which are in the interval $(0, 12)$ . Now we use the First Derivative Test. The sign of $P'(x)$ depends on the factors $(x-3)$ and $(x-10)$ . For $x < 3$ (e.g., $x=1$ ), $P'(1) = 6(1-3)(1-10) = 6(-2)(-9) > 0$ (increasing). For $3 < x < 10$ (e.g., $x=4$ ), $P'(4) = 6(4-3)(4-10) = 6(1)(-6) < 0$ (decreasing). For $x > 10$ (e.g., $x=11$ ), $P'(11) = 6(11-3)(11-10) = 6(8)(1) > 0$ (increasing). At $x=3$ , the function changes from increasing to decreasing, so it is a local maximum. At $x=10$ , the function changes from decreasing to increasing, so it is a local minimum. The question asks only for local maxima within the interval.
An online retailer's conversion rate function is $R(x) = x^2e^{-x}$ where $x$ represents the discount percentage (as a decimal). Given that $R'(x) = xe^{-x}(2-x)$ , a manager concludes that offering a 200% discount ( $x = 2$ ) maximizes conversion rate. What is wrong with this analysis?
Answer: A 200% discount is impossible in practice, so the manager should focus on the critical point behavior for $0 ≤ x ≤ 1$ While x=2 is indeed a critical point and the first derivative test confirms it's a local maximum (R'(1.5)>0, R'(2.5)<0), a 200% discount means giving customers twice the product value, which is economically impossible. The practical domain should be 0≤x≤1 (0% to 100% discount). Within this domain, we need to check x=0 and the endpoint x=1, comparing R(0)=0, R(1)= $e^{-1}$ ≈0.368. Choice A misses the practical constraint issue. Choice B ignores the domain restriction. Choice D questions the model validity unnecessarily.
The total profit function for a new software product is given by $P(x) = (x-30)^3 + 5000$ , where $x$ is the number of licenses sold (in hundreds), for $x \ge 0$ . Analyze the company's profit at the critical point $x=30$ .
Answer: Profit has neither a local maximum nor a local minimum at $x=30$ because profit is increasing on both sides of this point. First, we find the marginal profit (the first derivative) to identify critical points. $P'(x) = \frac{d}{dx}((x-30)^3 + 5000) = 3(x-30)^2$ . Setting $P'(x) = 0$ gives $3(x-30)^2 = 0$ , so $x=30$ is the only critical point. To classify this point, we use the First Derivative Test. We check the sign of $P'(x)$ on either side of $x=30$ . For $x < 30$ (e.g., $x=29$ ), $P'(29) = 3(29-30)^2 = 3(-1)^2 = 3 > 0$ . For $x > 30$ (e.g., $x=31$ ), $P'(31) = 3(31-30)^2 = 3(1)^2 = 3 > 0$ . Since $P'(x)$ is positive on both sides of $x=30$ , the profit function is increasing before and after this point. Therefore, $x=30$ is a critical point but it is not a local extremum (it is a stationary inflection point).
A marketing budget allocation function is $B(x) = x\ln(x) - 2x + 5$ where $x$ represents budget in thousands of dollars ( $x > 0$ ). Given $B'(x) = \ln(x) - 1$ , a financial analyst wants to use the first derivative test to optimize allocation. Which statement correctly interprets the results?
Answer: The critical point $x = e$ is a local minimum since $B'(1) = -1 < 0$ and $B'(10) = \ln(10) - 1 > 0$ When you encounter optimization problems in business calculus, you'll use the first derivative test to find and classify critical points. This requires finding where the derivative equals zero, then checking how the derivative's sign changes around those points. To find the critical point, set $B'(x) = \ln(x) - 1 = 0$ . Solving gives $\ln(x) = 1$ , so $x = e \approx 2.718$ . Now you must determine whether this is a maximum or minimum by testing the derivative's sign on either side of $x = e$ . The first derivative test works like this: if $B'(x)$ changes from negative to positive as you pass through a critical point, it's a local minimum. If it changes from positive to negative, it's a local maximum. Choice B correctly identifies $x = e$ as a local minimum by checking test points: $B'(1) = \ln(1) - 1 = 0 - 1 = -1 < 0$ (negative to the left of $e$ ) and $B'(10) = \ln(10) - 1 \approx 2.30 - 1 = 1.30 > 0$ (positive to the right of $e$ ). Since the derivative changes from negative to positive, we have a minimum. Choice A incorrectly calls it a maximum and misunderstands how the natural log's behavior relates to the critical point classification. Choice C is wrong because $\ln(x) = 1$ clearly has the solution $x = e$ . Choice D identifies the wrong critical point— $x = 1$ isn't even critical since $B'(1) = -1 \neq 0$ . Remember: always verify your critical points by substitution, and use concrete test values in the first derivative test rather than relying on intuition about function behavior.
The marginal revenue for a product is given by $R'(x) = (x-1)(x-3)e^{-x}$ for $x \ge 0$ . Determine the behavior of the total revenue function $R(x)$ on the interval between the critical points $x=1$ and $x=3$ .
Answer: Revenue is decreasing on the interval $(1, 3)$ . The behavior of the total revenue function $R(x)$ (whether it is increasing or decreasing) is determined by the sign of its derivative, the marginal revenue function $R'(x)$ . We are asked to analyze the behavior on the interval $(1, 3)$ . We can choose any test value within this interval, for example, $x=2$ . We evaluate the sign of $R'(x)$ at this point: $R'(2) = (2-1)(2-3)e^{-2} = (1)(-1)e^{-2} = -e^{-2}$ . Since $e^{-2}$ is positive, the value of $R'(2)$ is negative. As $R'(x)$ is continuous and only zero at $x=1$ and $x=3$ , it must be negative for the entire interval $(1, 3)$ . A negative derivative indicates that the original function is decreasing. Therefore, the total revenue function $R(x)$ is decreasing on the interval $(1, 3)$ .
The profit, $P(x)$ , from selling $x$ items is given by a function whose derivative is $P'(x) = x^2 e^{-0.1x}$ . How many local extrema does the profit function $P(x)$ have for $x>0$ ?
Answer: None, because $P'(x)$ is always positive for $x>0$ . Local extrema occur at critical points where the first derivative changes sign. First, find the critical points by setting $P'(x)=0$ . The equation is $x^2 e^{-0.1x} = 0$ . The term $e^{-0.1x}$ is always positive for any real $x$ . Therefore, the only way for $P'(x)$ to be zero is if $x^2 = 0$ , which means $x=0$ . However, the problem specifies the domain $x>0$ . In this domain, neither $x^2$ nor $e^{-0.1x}$ is ever zero, so there are no critical points for $x>0$ . Furthermore, for any $x>0$ , $x^2$ is positive and $e^{-0.1x}$ is positive, so their product $P'(x)$ is always positive. Since the derivative is always positive for $x>0$ , the profit function $P(x)$ is always increasing and has no local extrema in this domain.
A company's average cost function, $\bar{C}(x)$ , for producing $x$ units is $\bar{C}(x) = 2x + 15 + \frac{200}{x}$ for $x > 0$ . At what production level does the average cost change from decreasing to increasing?
Answer: At $x=10$ units, because this is a critical point where the average cost is minimized. To find where the average cost $\bar{C}(x)$ is increasing or decreasing, we need to find the sign of its derivative, $\bar{C}'(x)$ . First, find the derivative: $\bar{C}'(x) = \frac{d}{dx}(2x + 15 + 200x^{-1}) = 2 - 200x^{-2} = 2 - \frac{200}{x^2}$ . To find critical points, set $\bar{C}'(x) = 0$ . This gives $2 = \frac{200}{x^2}$ , which leads to $2x^2 = 200$ , so $x^2 = 100$ . Since $x > 0$ , the only critical point is $x=10$ . Now we use the First Derivative Test. For $0 < x < 10$ (e.g., $x=5$ ), $\bar{C}'(5) = 2 - \frac{200}{25} = 2 - 8 = -6 < 0$ , so the average cost is decreasing. For $x > 10$ (e.g., $x=20$ ), $\bar{C}'(20) = 2 - \frac{200}{400} = 2 - 0.5 = 1.5 > 0$ , so the average cost is increasing. Therefore, the average cost changes from decreasing to increasing at $x=10$ , indicating a local minimum.
The daily profit, $P(x)$ , of a company is given by $P(x) = kx^2 - x^3$ , where $x$ is the number of units produced and $k$ is a positive constant. If the company's profit has a local maximum at $x=100$ , what must be the value of the constant $k$ ?
Answer: $k=150$ When you encounter optimization problems involving profit functions, you need to find where the derivative equals zero and verify it's actually a maximum using the second derivative test. Given $P(x) = kx^2 - x^3$ , you first find the derivative: $P'(x) = 2kx - 3x^2$ . Since the profit has a local maximum at $x = 100$ , the derivative must equal zero at this point: $P'(100) = 0$ . Substituting: $2k(100) - 3(100)^2 = 0$ , which gives us $200k - 30,000 = 0$ . Solving for $k$ : $200k = 30,000$ , so $k = 150$ . You can verify this is indeed a maximum by checking that $P''(x) = 2k - 6x < 0$ when $x = 100$ . Let's examine why the other answers are incorrect. Answer (A) $k = 100$ would give $P'(100) = 200(100) - 30,000 = -10,000 \neq 0$ , so no critical point exists at $x = 100$ . Answer (B) $k = 300$ yields $P'(100) = 600(100) - 30,000 = 30,000 \neq 0$ , again failing the critical point condition. Answer (C) $k = 200$ produces $P'(100) = 400(100) - 30,000 = 10,000 \neq 0$ , also incorrect. Remember that optimization problems always require setting the first derivative equal to zero at the given critical point. Don't be tempted by "nice" round numbers like 100 or 200 – always work through the algebra systematically to find the exact value that satisfies the critical point condition.
A firm's cost function is $C(x) = \frac{250}{x} + x^2$ for $x>0$ . The firm's management wants to find the production level $x$ that minimizes cost. What is the correct conclusion from applying the First Derivative Test to this function?
Answer: Cost is minimized at $x=5$ , as the marginal cost changes from negative to positive at this point. To find the production level that minimizes cost, we first find the critical points of the cost function $C(x) = \frac{250}{x} + x^2$ . The marginal cost function is the derivative, $C'(x) = -\frac{250}{x^2} + 2x$ . We set the derivative to zero to find the critical points: $- \frac{250}{x^2} + 2x = 0$ . This implies $2x = \frac{250}{x^2}$ , which leads to $2x^3 = 250$ , so $x^3 = 125$ . The only real solution is $x=5$ . Now we use the First Derivative Test to classify this critical point. We test a value less than 5 (e.g., $x=1$ ): $C'(1) = -\frac{250}{1^2} + 2(1) = -250 + 2 = -248 < 0$ . We test a value greater than 5 (e.g., $x=10$ ): $C'(10) = -\frac{250}{10^2} + 2(10) = -2.5 + 20 = 17.5 > 0$ . Since the derivative changes from negative (decreasing cost) to positive (increasing cost) at $x=5$ , this point represents a local minimum for the cost function.

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