Continuity over an Interval - Calculus 1

1. Let $f(x) = \begin{cases} \frac{\sin(x)}{|x|} & x \ne 0 \\ a & x=0 \end{cases}$. Which statement best describes the continuity of $f(x)$ at $x=0$?
   Answer: The function has a jump discontinuity at $x=0$ for any value of $a$. To determine continuity at $x=0$, we must evaluate the left-hand and right-hand limits. The right-hand limit is $\lim_{x \to 0^+} \frac{\sin(x)}{|x|} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1$. The left-hand limit is $\lim_{x \to 0^-} \frac{\sin(x)}{|x|} = \lim_{x \to 0^-} \frac{\sin(x)}{-x} = -\lim_{x \to 0^-} \frac{\sin(x)}{x} = -1$. Since the left-hand limit ($-1$) does not equal the right-hand limit ($1$), the overall limit $\lim_{x \to 0} f(x)$ does not exist. Therefore, the function has a non-removable jump discontinuity at $x=0$, regardless of the value of $a$.
2. Suppose $f$ is a continuous function defined for all real numbers. If $f(x) = \frac{x^2+kx-6}{x-2}$ for $x \ne 2$, what is the value of $f(2)$?
   Answer: $5$ This problem tests your understanding of continuity and removable discontinuities. When a function is continuous everywhere but has a formula that's undefined at a point, you need to find what value makes the function continuous there. Since $f$ is continuous at $x = 2$ but the given formula has $x - 2$ in the denominator, there must be a removable discontinuity. This happens when the numerator also equals zero at $x = 2$, allowing you to cancel the problematic factor. For the numerator $x^2 + kx - 6$ to equal zero when $x = 2$: $2^2 + k(2) - 6 = 0$ $4 + 2k - 6 = 0$ $2k = 2$ $k = 1$ With $k = 1$, the function becomes $f(x) = \frac{x^2 + x - 6}{x - 2}$ for $x \neq 2$. Factor the numerator: $x^2 + x - 6 = (x + 3)(x - 2)$. Now you can cancel the $(x - 2)$ terms: $f(x) = \frac{(x + 3)(x - 2)}{x - 2} = x + 3$ for $x \neq 2$ Since $f$ is continuous at $x = 2$, we have $f(2) = 2 + 3 = 5$. Choice A is wrong because continuity gives us enough information to determine $f(2)$. Choice B gives the value when $k = -3$, and choice C gives the value when $k = -5$, but neither of these values of $k$ makes the function continuous at $x = 2$. Strategy tip: When you see a rational function that's supposed to be continuous at a point where the denominator is zero, immediately check what value in the numerator creates a removable discontinuity.
3. Let the function $f$ be defined by $f(x) = \begin{cases} ax^2 - 1 & \text{if } x < 2 \\ b & \text{if } x = 2 \\ x+a & \text{if } x > 2 \end{cases}$. If $f$ is continuous on the interval $[0, 4]$, what is the value of $b$?
   Answer: $3$ For $f$ to be continuous at $x=2$, the limit as $x$ approaches $2$ must exist and be equal to $f(2)$. This means the left-hand limit must equal the right-hand limit: $\lim_{x\to 2^-} f(x) = \lim_{x\to 2^+} f(x)$. Calculating the one-sided limits: Left-hand limit: $\lim_{x\to 2^-} (ax^2 - 1) = a(2^2) - 1 = 4a - 1$. Right-hand limit: $\lim_{x\to 2^+} (x+a) = 2+a$. Setting them equal: $4a - 1 = 2+a$, which simplifies to $3a = 3$, so $a=1$. For continuity, $f(2)$ must equal this limit. We have $f(2)=b$. The limit is $2+a = 2+1 = 3$. Therefore, $b=3$.
4. Let $g(x) = x^2 - 2x$ and $h(x) = x - 2$. Let $f(x) = \max(g(x), h(x))$. Which statement accurately describes the continuity of $f(x)$?
   Answer: $f(x)$ is continuous for all real numbers. Both $g(x)=x^2-2x$ and $h(x)=x-2$ are polynomials, so they are continuous for all real numbers. The function $f(x) = \max(g(x), h(x))$ takes the value of either $g(x)$ or $h(x)$, whichever is larger. A key theorem in calculus states that the maximum (and minimum) of two continuous functions is also a continuous function. To verify, we can check the points where the function might switch definitions, which is where $g(x)=h(x)$. $x^2 - 2x = x - 2 \implies x^2 - 3x + 2 = 0 \implies (x-1)(x-2)=0$. The transition points are $x=1$ and $x=2$. At $x=1$, $g(1)=-1$ and $h(1)=-1$. At $x=2$, $g(2)=0$ and $h(2)=0$. Since $g$ and $h$ are continuous and they meet at the transition points, the combined function $f(x)$ is continuous everywhere. The presence of sharp corners affects differentiability, not continuity.
5. The function $f(x) = \begin{cases} \frac{1-\cos(3x)}{x^2} & x \ne 0 \\ k & x=0 \end{cases}$ is continuous for all $x$. What is the value of $k$?
   Answer: $\frac{9}{2}$ For $f$ to be continuous at $x=0$, $k$ must be equal to $\lim_{x\to 0} f(x)$. We can evaluate this limit using L'Hôpital's Rule or by multiplying by the conjugate. Using the known limit $\lim_{u \to 0} \frac{1-\cos u}{u} = 0$ is insufficient. Let's use L'Hôpital's Rule: $\lim_{x\to 0} \frac{1-\cos(3x)}{x^2} = \lim_{x\to 0} \frac{3\sin(3x)}{2x}$. This is still $0/0$. Applying L'Hôpital's Rule again: $\lim_{x\to 0} \frac{9\cos(3x)}{2} = \frac{9\cos(0)}{2} = \frac{9}{2}$. Alternatively, using the identity $1-\cos(2\theta)=2\sin^2(\theta)$, we get $1-\cos(3x) = 2\sin^2(\frac{3x}{2})$. The limit becomes $\lim_{x\to 0} \frac{2\sin^2(\frac{3x}{2})}{x^2} = \lim_{x\to 0} 2 \left( \frac{\sin(\frac{3x}{2})}{x} \right)^2 = \lim_{x\to 0} 2 \left( \frac{3}{2} \frac{\sin(\frac{3x}{2})}{\frac{3x}{2}} \right)^2 = 2 \left( \frac{3}{2} \cdot 1 \right)^2 = 2 \left( \frac{9}{4} \right) = \frac{9}{2}$.
6. Let $f(x) = e^x$ and $g(x) = \cos(x)$. The Intermediate Value Theorem guarantees that the equation $f(x)=2g(x)$ has a solution on which of the following intervals?
   Answer: $[0, \frac{\pi}{2}]$ To apply the IVT, we want to find a root of the function $h(x) = f(x) - 2g(x) = e^x - 2\cos(x)$. This function is continuous everywhere. We need to find an interval $[a, b]$ where $h(a)$ and $h(b)$ have opposite signs. Let's test the intervals. A) $h(-\pi) = e^{-\pi} - 2\cos(-\pi) = e^{-\pi} - 2(-1) = e^{-\pi}+2 > 0$. $h(-\frac{\pi}{2}) = e^{-\pi/2} - 2\cos(-\frac{\pi}{2}) = e^{-\pi/2} - 0 > 0$. No guarantee. B) $h(-\frac{\pi}{2}) > 0$. $h(0) = e^0 - 2\cos(0) = 1 - 2(1) = -1 < 0$. The sign changes from positive to negative. However, this is not an option. Let me re-check. No, this interval isn't an option. Let's check C. $h(0) = -1 < 0$. $h(\frac{\pi}{2}) = e^{\pi/2} - 2\cos(\frac{\pi}{2}) = e^{\pi/2} - 0 > 0$. The sign changes from negative to positive on $[0, \frac{\pi}{2}]$, so the IVT guarantees a solution in this interval. D) $h(\frac{\pi}{2}) > 0$. $h(\pi) = e^{\pi} - 2\cos(\pi) = e^{\pi} - 2(-1) = e^{\pi}+2 > 0$. No guarantee.
7. Let $f(x) = \frac{1}{x-2}$ and $g(x) = \sqrt{x}$. The composite function $h(x) = f(g(x))$ is continuous on which of the following sets?
   Answer: $\{x \in \mathbb{R} \mid x \ge 0 \text{ and } x \ne 4\}$ The composite function is $h(x) = f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x}-2}$. For $h(x)$ to be continuous, it must be defined. Two conditions must be met: 1. The inner function, $g(x)=\sqrt{x}$, must be defined. This requires $x \ge 0$. 2. The denominator of the resulting function cannot be zero. This means $\sqrt{x}-2 \ne 0$, which implies $\sqrt{x} \ne 2$, so $x \ne 4$. Combining both conditions, the set of all points where $h(x)$ is defined and continuous is all real numbers $x$ such that $x \ge 0$ and $x \ne 4$.
8. Let $f(x) = \begin{cases} \sqrt{1-x} & \text{if } x \le 1 \\ \ln(x) & \text{if } x > 1 \end{cases}$. Which of the following is the largest interval on which $f(x)$ is continuous?
   Answer: $(-\infty, \infty)$ First, check the continuity of each piece on its given interval. The function $\sqrt{1-x}$ is continuous for $1-x \ge 0$, which is $x \le 1$. The function $\ln(x)$ is continuous for $x > 0$, so it is certainly continuous on its given interval $x > 1$. Next, check for continuity at the breakpoint, $x=1$. We need the one-sided limits to equal the function value. $f(1) = \sqrt{1-1} = 0$. $\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} \sqrt{1-x} = \sqrt{1-1} = 0$. $\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} \ln(x) = \ln(1) = 0$. Since the left-hand limit, right-hand limit, and the function value are all equal to $0$ at $x=1$, the function is continuous at $x=1$. Because the pieces are continuous on their intervals and the function is continuous at the breakpoint, $f(x)$ is continuous for all real numbers. Thus, the largest interval of continuity is $(-\infty, \infty)$.
9. On which of the following closed intervals is the function $h(x) = \ln(4-x^2)$ continuous?
   Answer: $[-1, \sqrt{3}]$ The function $h(x) = \ln(u)$ is continuous wherever $u > 0$. For $h(x) = \ln(4-x^2)$, we need $4-x^2 > 0$, which means $x^2 < 4$. The solution to this inequality is $-2 < x < 2$. So, $h(x)$ is continuous on the open interval $(-2, 2)$. A function is continuous on a closed interval $[a, b]$ if $[a, b]$ is a subset of the domain of continuity. We must check which of the given closed intervals is a subset of $(-2, 2)$. A) $[-2, 2]$ is not a subset because the endpoints $-2$ and $2$ are not in $(-2, 2)$. B) Since $\sqrt{3} \approx 1.732$, the interval $[-1, \sqrt{3}]$ is a subset of $(-2, 2)$. C) $[0, 3]$ is not a subset because it includes values greater than $2$. D) $[-\sqrt{5}, -1]$ is not a subset because $-\sqrt{5} \approx -2.236$, which is less than $-2$.
10. Let $f(x) = \frac{x^2 - 4}{x^3 - 2x^2 - 8x}$. On which of the following intervals is $f$ continuous?
    Answer: $(-1, 3)$ To find points of discontinuity, we factor the numerator and the denominator. Numerator: $x^2 - 4 = (x-2)(x+2)$. Denominator: $x^3 - 2x^2 - 8x = x(x^2 - 2x - 8) = x(x-4)(x+2)$. So, $f(x) = \frac{(x-2)(x+2)}{x(x-4)(x+2)}$. The function is undefined, and therefore not continuous, at $x=0$, $x=4$, and $x=-2$. At $x=-2$, the discontinuity is removable since the factor $(x+2)$ cancels. At $x=0$ and $x=4$, the discontinuities are non-removable (vertical asymptotes). A function is continuous on an interval if it is continuous at every point in that interval. We must find which option does not contain any of the points $x=-2, 0, 4$. A) $[-3, 0)$ contains $x=-2$. B) $(-1, 3)$ does not contain $-2, 0$, or $4$. C) $(-4, 4)$ contains $x=-2$ and $x=0$. D) $[1, 5]$ contains $x=4$.
11. A function $f$ is continuous on $[a, b]$, and the Extreme Value Theorem guarantees that it attains a maximum value $M$ and a minimum value $m$ on $[a, b]$. What does the Intermediate Value Theorem add to our knowledge about the function's range?
    Answer: It guarantees that the function's range is exactly the interval $[m, M]$. When you encounter questions about the Extreme Value Theorem and Intermediate Value Theorem working together, focus on how these two fundamental theorems combine to give you complete information about a continuous function's behavior on a closed interval. The Extreme Value Theorem tells you that $f$ achieves both a maximum value $M$ and minimum value $m$ somewhere on $[a,b]$. But this alone doesn't tell you whether $f$ takes on every value between $m$ and $M$. The Intermediate Value Theorem fills this gap: since $f$ is continuous and takes on values $m$ and $M$, it must take on every value between them. Therefore, the range is exactly $[m,M]$. Choice A is wrong because the Intermediate Value Theorem only guarantees a root if $f(a)$ and $f(b)$ have opposite signs. This isn't given in the problem. Choice B incorrectly connects the condition $f(a) \neq f(b)$ to whether $m < M$. Even if $f(a) = f(b)$, the function could still vary in between, making $m < M$. Choice C is incorrect because continuous functions on closed intervals can definitely fail to be one-to-one. For example, $f(x) = x^2$ on $[-1,1]$ satisfies all conditions but isn't one-to-one. Remember this key insight: the Extreme Value Theorem gives you the "endpoints" of the range ($m$ and $M$), while the Intermediate Value Theorem guarantees you get everything in between. Together, they completely determine the range as $[m,M]$.
12. Let $f(x) = (1+3x)^{1/x}$ for $x \ne 0$. If $f$ is defined to be continuous at $x=0$, what must be the value of $f(0)$?
    Answer: $e^3$ For $f$ to be continuous at $x=0$, $f(0)$ must equal $\lim_{x\to 0} (1+3x)^{1/x}$. This limit is of the indeterminate form $1^\infty$. We can solve this by letting $y = (1+3x)^{1/x}$ and finding the limit of $\ln(y)$. $\lim_{x\to 0} \ln(y) = \lim_{x\to 0} \frac{\ln(1+3x)}{x}$. Using L'Hôpital's Rule, this is $\lim_{x\to 0} \frac{3/(1+3x)}{1} = 3$. Since $\lim_{x\to 0} \ln(y) = 3$, we have $\lim_{x\to 0} y = e^3$. Alternatively, this is a direct application of the known limit form $\lim_{x\to 0} (1+ax)^{1/x} = e^a$, with $a=3$. Thus, $f(0)$ must be $e^3$.
13. Let $f(x) = \frac{|x-a|}{x^2-a^2}$ where $a \ne 0$. Which statement correctly describes the discontinuities of $f(x)$?
    Answer: A jump discontinuity at $x=a$ and an infinite discontinuity at $x=-a$. The denominator is $x^2-a^2 = (x-a)(x+a)$, so there are discontinuities at $x=a$ and $x=-a$. At $x=-a$, the numerator is $|-a-a|=|-2a|=2|a|$, which is non-zero. Since the numerator is non-zero and the denominator is zero, there is an infinite discontinuity at $x=-a$. Now consider $x=a$. We need to check the one-sided limits. For $x \to a^+$, $x-a > 0$, so $|x-a|=x-a$. The limit is $\lim_{x\to a^+} \frac{x-a}{(x-a)(x+a)} = \lim_{x\to a^+} \frac{1}{x+a} = \frac{1}{2a}$. For $x \to a^-$, $x-a < 0$, so $|x-a|=-(x-a)$. The limit is $\lim_{x\to a^-} \frac{-(x-a)}{(x-a)(x+a)} = \lim_{x\to a^-} \frac{-1}{x+a} = \frac{-1}{2a}$. Since the one-sided limits exist but are not equal, there is a jump discontinuity at $x=a$.
14. For what values of the constants $a$ and $b$ is the function $f(x)$ continuous on $(-\infty, \infty)$? $$f(x) = \begin{cases} ax-b & x \le -1 \\ 2x^2+3ax+b & -1 < x \le 1 \\ 4 & x > 1 \end{cases}$$
    Answer: $a = \frac{3}{4}, b = -\frac{1}{4}$ For continuity at $x=-1$, the left-hand limit must equal the right-hand limit: $a(-1)-b = 2(-1)^2+3a(-1)+b$, which simplifies to $-a-b = 2-3a+b$, or $2a-2b=2$, so $a-b=1$. For continuity at $x=1$, the left-hand limit must equal the right-hand limit: $2(1)^2+3a(1)+b = 4$, which simplifies to $2+3a+b=4$, or $3a+b=2$. We now have a system of two linear equations: $a-b=1$ and $3a+b=2$. Adding the two equations gives $4a=3$, so $a=\frac{3}{4}$. Substituting this into $a-b=1$ gives $\frac{3}{4}-b=1$, so $b = \frac{3}{4}-1 = -\frac{1}{4}$.
15. Find the largest value $b$ such that the function $f(x) = \ln(x^2-x-12)$ is continuous on the interval $(-\infty, b)$.
    Answer: $-3$ The natural logarithm function $\ln(u)$ is continuous for $u > 0$. Thus, we need $x^2-x-12 > 0$. Factoring the quadratic, we get $(x-4)(x+3) > 0$. The roots are at $x=4$ and $x=-3$. The parabola opens upwards, so the expression is positive when $x > 4$ or $x < -3$. The domain of $f(x)$ is $(-\infty, -3) \cup (4, \infty)$. The function is continuous on these two disjoint intervals. We are looking for the largest value $b$ such that the interval $(-\infty, b)$ is a subset of the domain. This corresponds to the interval $(-\infty, -3)$. Thus, the largest possible value for $b$ is $-3$.
16. Let $f(x)$ and $g(x)$ be functions that are both continuous on the interval $[-5, 5]$. Suppose $f(3)=4$ and $g(x) = x^2 - 13$. Which of the following functions is NOT necessarily continuous on the interval $[-5, 5]$?
    Answer: $h(x) = \frac{f(x)}{g(x)}$ We use the properties of continuous functions. A) The sum of two functions continuous on an interval is also continuous on that interval. So, $h(x)=f(x)+g(x)$ is continuous on $[-5, 5]$. B) The product of two functions continuous on an interval is also continuous on that interval. So, $h(x)=f(x)g(x)$ is continuous on $[-5, 5]$. C) The quotient of two continuous functions, $f/g$, is continuous wherever the denominator $g(x)$ is not zero. We must check if $g(x)=0$ for any $x$ in $[-5, 5]$. Set $g(x) = x^2 - 13 = 0$. This gives $x^2 = 13$, so $x = \pm\sqrt{13}$. Since $3 < \sqrt{13} < 4$, both $\sqrt{13}$ and $-\sqrt{13}$ are in the interval $[-5, 5]$. Because the denominator is zero at these points, $h(x) = f(x)/g(x)$ is not continuous on the entire interval $[-5, 5]$. D) The composition of continuous functions is continuous. Since $f(x)$ is continuous on $[-5, 5]$ and $g(x)$ is continuous everywhere, the composite function $h(x)=g(f(x))$ is continuous on $[-5, 5]$. The value $f(3)=4$ is irrelevant information.
17. For what value of the constant $c$ is the function $f(x) = \begin{cases} c x^2 + 2x & x < 2 \\ x^3 - cx & x \ge 2 \end{cases}$ continuous on $(-\infty, \infty)$?
    Answer: $\frac{2}{3}$ For the function to be continuous, the limit from the left and the limit from the right must be equal at the point $x=2$. We set the values of the two pieces equal at $x=2$: $\lim_{x\to 2^-} (c x^2 + 2x) = \lim_{x\to 2^+} (x^3 - cx)$. Plugging in $x=2$, we get $c(2^2) + 2(2) = (2)^3 - c(2)$. This simplifies to $4c + 4 = 8 - 2c$. Adding $2c$ to both sides gives $6c + 4 = 8$. Subtracting 4 from both sides gives $6c = 4$. Therefore, $c = \frac{4}{6} = \frac{2}{3}$.
18. Consider the function $f(x) = \sec(x)$ on the interval $[0, \pi]$. Which statement is true?
    Answer: $f$ is continuous on $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$. The function $f(x) = \sec(x) = \frac{1}{\cos(x)}$ is discontinuous where $\cos(x) = 0$. On the interval $[0, \pi]$, this occurs at $x=\frac{\pi}{2}$. At this point, the limit from the left is $\lim_{x \to (\pi/2)^-} \sec(x) = +\infty$ and the limit from the right is $\lim_{x \to (\pi/2)^+} \sec(x) = -\infty$. This is an infinite discontinuity. The function is continuous on the intervals where it is defined within $[0, \pi]$, which are $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$.
19. On which of the following intervals is the function $f(x) = \frac{\csc(x)}{2\cos(x) - 1}$ continuous?
    Answer: $(\pi, \frac{5\pi}{3})$ The function can be written as $f(x) = \frac{1}{\sin(x)(2\cos(x) - 1)}$. Discontinuities occur when the denominator is zero. This happens if $\sin(x)=0$ or $2\cos(x)-1=0$. 1. $\sin(x)=0$ when $x=n\pi$ for any integer $n$. In the range $[0, 2\pi]$, this occurs at $x=0, \pi, 2\pi$. 2. $2\cos(x)-1=0$ when $\cos(x)=1/2$. In the range $[0, 2\pi]$, this occurs at $x=\pi/3$ and $x=5\pi/3$. The points of discontinuity are $0, \pi/3, \pi, 5\pi/3, 2\pi$, etc. We need to find an interval from the options that avoids all these points. A) $(\pi/4, 3\pi/4)$ contains $\pi/3$. B) $(\pi/3, 5\pi/3)$ contains $\pi$. C) $(\pi, 5\pi/3)$ is the interval between two consecutive discontinuities, so $f(x)$ is continuous here. D) $(0, \pi)$ contains $\pi/3$.
20. On which of the following intervals is the function $f(x) = \frac{\sqrt{x+3}}{x^2-x-2}$ continuous?
    Answer: $[-3, -1) \cup (-1, 2) \cup (2, \infty)$ The function's continuity is determined by its domain. There are two conditions for the domain. First, the term inside the square root must be non-negative: $x+3 \ge 0$, which implies $x \ge -3$. Second, the denominator cannot be zero: $x^2-x-2 \ne 0$. Factoring the quadratic gives $(x-2)(x+1) \ne 0$, so $x \ne 2$ and $x \ne -1$. We must satisfy all conditions. We need $x \ge -3$ but $x \ne -1$ and $x \ne 2$. Combining these gives the set of intervals $[-3, -1) \cup (-1, 2) \cup (2, \infty)$. The function is continuous on this entire set.

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