Differentiability vs. Continuity - Business Calculus

1. The total cost C(x) in dollars for manufacturing x units of a certain product is given by the piecewise function: $C(x) = \begin{cases} 5x^2 + 100 & \text{if } 0 \le x \le 10 \\ 40x + 200 & \text{if } x > 10 \end{cases}$ Which of the following statements accurately describes the cost function at the production level of x=10 units?
   Answer: The cost function is continuous but not differentiable at x=10. To check for continuity at x=10, we evaluate the function value and the limit from both sides. C(10) = 5(10)^2 + 100 = 500 + 100 = 600. The limit from the left is \lim_{x \to 10^-} (5x^2 + 100) = 600. The limit from the right is \lim_{x \to 10^+} (40x + 200) = 40(10) + 200 = 600. Since the left-hand limit, right-hand limit, and function value are all equal, the function is continuous at x=10. To check for differentiability, we find the derivatives of the pieces. For x < 10, C'(x) = 10x. The left-hand derivative at x=10 is 10(10) = 100. For x > 10, C'(x) = 40. The right-hand derivative at x=10 is 40. Since the left-hand derivative (100) does not equal the right-hand derivative (40), the function is not differentiable at x=10.
2. The function $g(x) = |x^2 - 4|$ is being analyzed for continuity and differentiability. At which point(s) is $g(x)$ continuous but not differentiable?
   Answer: At $x = -2$ and $x = 2$ because the absolute value creates sharp corners When analyzing functions involving absolute values, you need to examine both continuity and differentiability carefully. Continuity requires the function to have no breaks or jumps, while differentiability requires the function to have no sharp corners or cusps. For $g(x) = |x^2 - 4|$, start by finding where the expression inside the absolute value equals zero: $x^2 - 4 = 0$, which gives $x = -2$ and $x = 2$. These are critical points where the absolute value function might create sharp corners. At these points, $g(x)$ is continuous because the function value approaches the same limit from both sides. However, the derivative changes abruptly at these points. When $x^2 - 4 < 0$ (between -2 and 2), $g(x) = -(x^2 - 4) = 4 - x^2$, so $g'(x) = -2x$. When $x^2 - 4 > 0$ (outside the interval [-2,2]), $g(x) = x^2 - 4$, so $g'(x) = 2x$. At $x = 2$, the left derivative is $-4$ while the right derivative is $+4$, creating a sharp corner where the function isn't differentiable. Option A is wrong because absolute values can indeed create non-differentiable points. Option B incorrectly identifies $x = 0$ as the problem point, but the function is smooth there. Option C correctly identifies the points but gives the wrong reason—it's not about the derivative changing sign, but about the derivative being undefined due to sharp corners. Option D correctly identifies both the points and the geometric reason. Remember: absolute value functions create sharp corners at zeros of the inner expression, making them non-differentiable at those specific points.
3. Which of the following functions provides a counterexample to the assertion that 'If a function is continuous for all real numbers, then it must be differentiable for all real numbers'?
   Answer: f(x) = (x-2)^{2/3} A counterexample must satisfy the 'if' condition (be continuous for all real numbers) but fail the 'then' condition (not be differentiable for all real numbers). - A and B are both continuous and differentiable for all real numbers, so they are not counterexamples. - D is not continuous at x=0 because \lim_{x \to 0} f(x) = 0 but f(0)=1. Since it does not satisfy the 'if' condition, it cannot be a counterexample. - C, f(x) = (x-2)^{2/3}, is continuous for all real numbers. However, its derivative is f'(x) = \frac{2}{3}(x-2)^{-1/3} = \frac{2}{3\sqrt[3]{x-2}}. This derivative is undefined at x=2, where the graph has a cusp. Therefore, f(x) is continuous everywhere but not differentiable at x=2, making it a valid counterexample.
4. Consider the following statements about a function f at a point c: I. f is continuous at c. II. f is differentiable at c. Which statement correctly describes the logical relationship between these two properties?
   Answer: Statement I is a necessary condition for Statement II, but not a sufficient one. The relationship is that differentiability implies continuity. This can be rephrased in terms of necessary and sufficient conditions. If f is differentiable at c (II), then it must be continuous at c (I). This means continuity (I) is a necessary condition for differentiability (II). You cannot have differentiability without continuity. However, if f is continuous at c (I), it is not guaranteed to be differentiable at c (II). For example, f(x)=|x| is continuous at x=0 but not differentiable. This means continuity (I) is not a sufficient condition for differentiability (II). Therefore, Statement I is necessary, but not sufficient, for Statement II.
5. Consider the piecewise function $f(x) = \begin{cases} ax^3 + bx & \text{if } x \leq 1 \\ cx^2 + dx + e & \text{if } x > 1 \end{cases}$. If $f(x)$ is continuous at $x = 1$ but not differentiable there, and $f(1) = 5$, which condition must be satisfied?
   Answer: $a + b = c + d + e = 5$ and left/right derivatives are unequal For continuity at $x = 1$: the left limit $a + b$, right limit $c + d + e$, and function value must all equal 5. For non-differentiability, the left and right derivatives must be unequal. The left derivative is $3a + b$ and the right derivative is $2c + d$, so we need $3a + b \neq 2c + d$.
6. The monthly profit P (in dollars) of a company is modeled by the function P(t) = 1000|t - 6| + 5000, where t is the number of months since the start of the year (0 \le t \le 12). Which statement best describes the profit function at t=6?
   Answer: The profit function has a sharp corner at t=6, meaning it is continuous but not differentiable. The function P(t) = 1000|t - 6| + 5000 involves an absolute value. The absolute value function |x| is continuous everywhere but not differentiable at x=0 (where it has a corner). In this model, the argument of the absolute value is t-6, which is zero when t=6. Therefore, the function P(t) will have a sharp corner at t=6. A function with a corner is continuous but not differentiable at that point. At t=6, P(6) = 1000|0| + 5000 = 5000, and the function approaches this value from both sides, so it is continuous. However, the rate of change abruptly shifts from -1000 (for t<6) to +1000 (for t>6), so it is not differentiable.
7. Let f(x) be a function for which we know that f(3) = 5, \lim_{x \to 3} f(x) = 5, and the derivative f'(x) is undefined at x=3. Which of the following statements must be true?
   Answer: The graph of f(x) could have a corner or a cusp at x=3 while remaining continuous. When you encounter questions about derivatives being undefined, it's crucial to understand that there are several reasons why $f'(x)$ might not exist at a point, and they don't all imply discontinuity. The key insight here is recognizing what continuity requires versus what differentiability requires. Since $f(3) = 5$ and $\lim_{x \to 3} f(x) = 5$, the function is indeed continuous at $x = 3$ because the function value equals the limit. However, a function can be continuous at a point while still having an undefined derivative there. Answer D is correct because when $f'(3)$ is undefined but the function remains continuous, the graph typically has a sharp point—either a corner (where left and right derivatives exist but are different) or a cusp (where one or both one-sided derivatives are infinite). Think of $y = |x|$ at $x = 0$: continuous but not differentiable due to the sharp corner. Answer A is wrong because vertical asymptotes occur when the function itself approaches infinity, not when just the derivative is undefined. Answer B incorrectly assumes that an undefined derivative always means discontinuity—this is a common misconception. Differentiability requires continuity, but continuity doesn't guarantee differentiability. Answer C is false because the given conditions are perfectly compatible, as demonstrated by functions like $f(x) = |x - 3| + 5$. Remember: a function can be continuous everywhere but differentiable nowhere, or continuous at a point but not differentiable there. Continuity and differentiability are related but distinct concepts.
8. Let $F(x)$ be defined by $x^2 + y^2 = 25$ where $y = F(x)$ and $y \geq 0$. The function $F(x) = \sqrt{25 - x^2}$ represents the upper semicircle. At which point(s) is $F(x)$ continuous but not differentiable?
   Answer: At $x = 5$ and $x = -5$ because the tangent line is vertical there For $F(x) = \sqrt{25 - x^2}$, using the chain rule gives $F'(x) = \frac{-x}{\sqrt{25-x^2}}$. At $x = \pm 5$, the denominator approaches 0, making the slope infinite (vertical tangent). The function is continuous at these endpoints but not differentiable due to the vertical tangent lines.
9. A function f(x) is differentiable at x=c. Which of the following statements must be true as a consequence?
   Answer: The limit \lim_{x \to c} f(x) must exist and be equal to f(c). The fundamental theorem relating differentiability and continuity states that if a function f is differentiable at a point c, it must be continuous at c. The definition of continuity at c is that \lim_{x \to c} f(x) = f(c). Therefore, statement B must be true. - A is incorrect; f'(c) could be any real number, not necessarily 0 (which defines a horizontal tangent). - C is incorrect; a function can be differentiable at a single point without being differentiable in the surrounding interval. For example, f(x) = x^2 if x is rational and f(x) = 0 if x is irrational is differentiable only at x=0. - D is incorrect; this is the definition of the continuity of the derivative f'(x), not a necessary condition for the differentiability of f(x) itself. A function can be differentiable at a point without its derivative being continuous at that point.
10. The number of active users N(t) for a new app t days after launch is modeled by N(t) = 100t^{1/3}. Which of the following statements best describes the model at the moment of launch, t=0?
    Answer: The function is continuous at t=0 but has a vertical tangent, meaning it is not differentiable. When analyzing functions at specific points, you need to check both continuity and differentiability. For $N(t) = 100t^{1/3}$, let's examine what happens at launch ($t = 0$). First, check continuity at $t = 0$. Since $N(0) = 100(0)^{1/3} = 0$, and the limit as $t$ approaches 0 from the right is also 0, the function is continuous at $t = 0$. This makes business sense—you start with zero users at launch. Next, examine differentiability by finding the derivative: $N'(t) = 100 \cdot \frac{1}{3}t^{-2/3} = \frac{100}{3t^{2/3}}$. At $t = 0$, this becomes $\frac{100}{3(0)^{2/3}} = \frac{100}{0}$, which is undefined (approaches infinity). This creates a vertical tangent line at $t = 0$, meaning the function is not differentiable there. Answer A is incorrect because while the function is continuous, it doesn't have a horizontal tangent—it has a vertical one. Answer B is wrong because the function is continuous, not discontinuous. Answer C is false because the function isn't differentiable at $t = 0$; the growth rate is actually infinite (vertical tangent), not 100 users/day. Answer D correctly identifies that the function is continuous but has a vertical tangent, making it non-differentiable at $t = 0$. Study tip: Functions with fractional exponents between 0 and 1 (like $t^{1/3}$) often create vertical tangents at $t = 0$. Always check both continuity and differentiability separately—one doesn't guarantee the other.
11. A company's profit function $P(x)$ (in thousands of dollars) is continuous for all production levels $x \geq 0$. At $x = 100$ units, the right-hand derivative is $P'_+(100) = 3$ and the left-hand derivative is $P'_-(100) = 3$. However, $P'(x)$ does not exist at several points near $x = 100$. What can be concluded about $P(x)$ at $x = 100$?
    Answer: $P(x)$ is differentiable at $x = 100$ and the marginal profit is $3000 per unit Differentiability at a point depends only on the behavior at that specific point, not on nearby points. Since the left and right derivatives both exist and are equal at $x = 100$, the function is differentiable there with $P'(100) = 3$ (representing $3000 per unit since profit is in thousands). The fact that the derivative doesn't exist at nearby points is irrelevant to differentiability at $x = 100$ itself.
12. Let the function f(x) be defined as: $f(x) = \begin{cases} ax^2 + b & \text{if } x < 1 \\ 2x - 3 & \text{if } x \ge 1 \end{cases}$ For what values of the constants a and b is the function f(x) differentiable at x=1?
    Answer: a = 1, b = -2 For f(x) to be differentiable at x=1, it must first be continuous at x=1. This means the limits from both sides must be equal: \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) a(1)^2 + b = 2(1) - 3 a + b = -1 (Equation 1) Second, the derivatives from both sides must be equal at x=1. The derivative is: f'(x) = \begin{cases} 2ax & \text{if } x < 1 \\ 2 & \text{if } x > 1 \end{cases} Setting the left-hand derivative equal to the right-hand derivative at x=1: 2a(1) = 2 2a = 2, which means a = 1. Now, substitute a=1 into Equation 1: 1 + b = -1 b = -2. Thus, the function is differentiable at x=1 when a = 1 and b = -2.
13. The revenue R(x) from selling x units of a product is continuous for x \ge 0. The marginal revenue function R'(x) is also continuous, except at x=500, where \lim_{x \to 500^-} R'(x) = 10 and \lim_{x \to 500^+} R'(x) = 8. What does this imply about the revenue function R(x) at x=500?
    Answer: R(x) is continuous but not differentiable at x=500. The problem states that the revenue function R(x) is continuous for x \ge 0, so it is continuous at x=500. To be differentiable at x=500, the left-hand derivative must equal the right-hand derivative. The given limits of the marginal revenue function, \lim_{x \to 500^-} R'(x) = 10 and \lim_{x \to 500^+} R'(x) = 8, represent the left-hand and right-hand derivatives, respectively. Since 10 \ne 8, the derivative R'(500) does not exist. Therefore, the function R(x) is continuous but not differentiable at x=500. The graph of R(x) would have a corner at this point.
14. Suppose f(x) is a function such that \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} does not exist. Which of the following statements can be concluded with certainty?
    Answer: f(x) is not differentiable at x=c. The expression \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} is the definition of the derivative of f(x) at the point x=c, denoted f'(c). The problem states that this limit does not exist. Therefore, by definition, the function f(x) is not differentiable at x=c. - A and C might be true, as discontinuity implies non-differentiability, but they are not certain. A function can be continuous at a point but not differentiable there (e.g., f(x) = |x| at c=0). In that case, f(x) is continuous and \lim_{x \to c} f(x) exists, but the function is still not differentiable. - D is incorrect. For the limit expression to be written, f(c) must be defined.
15. Consider $h(x) = x^{2/3}$ at $x = 0$. A student claims that since $\lim_{x \to 0} h(x) = h(0) = 0$, the function is continuous, and since power functions are differentiable, $h(x)$ must be differentiable at $x = 0$. What is wrong with this reasoning?
    Answer: The derivative $\frac{d}{dx}[x^{2/3}] = \frac{2}{3}x^{-1/3}$ is undefined at $x = 0$ The student correctly identifies that $h(x) = x^{2/3}$ is continuous at $x = 0$. However, when differentiating: $h'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$. At $x = 0$, this expression involves division by zero, so the derivative is undefined. While power functions are generally differentiable, those with exponents between 0 and 1 have vertical tangent lines at $x = 0$, making them non-differentiable there despite being continuous.

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