Differentiability & Continuity - Calculus 1

1. For what value of $k$ is the function $f(x) = \begin{cases} x^2 - 1 & x < 2 \\ k(x-2) + 3 & x \ge 2 \end{cases}$ differentiable at $x=2$?
   Answer: $k=4$ First, the function must be continuous at $x=2$. This means the limits from both sides must be equal to the function value. $\lim_{x \to 2^-} (x^2 - 1) = 2^2 - 1 = 3$. The value from the right is $f(2) = k(2-2) + 3 = 3$. Since both equal 3, the function is continuous for any value of $k$. Next, the derivatives must match. The derivative of $x^2 - 1$ is $2x$. At $x=2$, the left-hand derivative is $2(2) = 4$. The derivative of $k(x-2)+3 = kx - 2k + 3$ is $k$. For the function to be differentiable, we must set the derivatives equal: $k=4$.
2. Let $f(x) = \frac{|x-1|(x+2)}{x-1}$. At which value(s) of $x$ is the function not differentiable?
   Answer: $x=1$ only The function is not defined at $x=1$ because it causes division by zero. A function cannot be differentiable at a point where it is not defined. We can analyze the function by rewriting it as a piecewise function: $$f(x) = \begin{cases} \frac{(x-1)(x+2)}{x-1} = x+2 & x > 1 \\ \frac{-(x-1)(x+2)}{x-1} = -(x+2) & x < 1 \end{cases}$$ As $x \to 1^+$, $f(x) \to 3$. As $x \to 1^-$, $f(x) \to -3$. This shows a jump discontinuity at $x=1$. Since the function is not continuous at $x=1$, it is not differentiable at $x=1$. For all other values of $x$, the function is linear and therefore differentiable.
3. Let $f$ be a function such that $\lim_{x \to -1} f(x) = 4$ but $f(-1)$ is undefined. Which of the following statements must be true?
   Answer: $f$ is not continuous at $x=-1$ and therefore not differentiable at $x=-1$. For a function to be continuous at a point $c$, three conditions must be met: $f(c)$ must be defined, $\lim_{x \to c} f(x)$ must exist, and these two values must be equal. Here, $f(-1)$ is undefined, so the first condition for continuity fails. A function cannot be differentiable at a point where it is not continuous. Therefore, $f$ is not continuous at $x=-1$ and, as a consequence, cannot be differentiable at $x=-1$.
4. A function $f$ has a derivative given by $f'(x) = \frac{1}{x-3}$. What can be concluded about the differentiability of the original function $f(x)$ at $x=3$?
   Answer: $f(x)$ is not differentiable at $x=3$ because $f'(x)$ is undefined. The derivative $f'(x)$ is given as $\frac{1}{x-3}$. At $x=3$, the expression for $f'(x)$ is undefined due to division by zero. For $f(x)$ to be differentiable at $x=3$, $f'(3)$ would need to be a finite value. Since $f'(3)$ does not exist, the function $f(x)$ is not differentiable at $x=3$. The original function, which would be $\ln|x-3| + C$, is not even continuous at $x=3$ (it has a vertical asymptote), which is a more fundamental reason for its non-differentiability.
5. The function $f(x) = |\cos(x)|$ is not differentiable at which of the following x-values?
   Answer: $x = \pi/2$ The function $f(x) = |g(x)|$ is typically not differentiable where $g(x)=0$. In this case, $g(x)=\cos(x)$. We need to find where $\cos(x)=0$. This occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. At these points, the graph of $|\cos(x)|$ has sharp corners. Let's check $x=\pi/2$. The derivative of $\cos(x)$ is $-\sin(x)$. For $x$ slightly less than $\pi/2$, $\cos(x)>0$, so $f(x)=\cos(x)$ and the slope is $-\sin(\pi/2)=-1$. For $x$ slightly greater than $\pi/2$, $\cos(x)<0$, so $f(x)=-\cos(x)$ and the slope is $\sin(\pi/2)=1$. Since $-1 \neq 1$, the function is not differentiable at $x=\pi/2$.
6. Let $f(x) = \begin{cases} x+2 & x<0 \\ (x+1)^2 & x \ge 0 \end{cases}$. Which statement accurately describes $f(x)$ at $x=0$?
   Answer: It is neither continuous nor differentiable. First, check for continuity at $x=0$. The limit from the left is $\lim_{x \to 0^-} (x+2) = 2$. The limit from the right is $\lim_{x \to 0^+} (x+1)^2 = (0+1)^2 = 1$. Since the left and right limits are not equal ($2 \neq 1$), the function has a jump discontinuity at $x=0$. Because the function is not continuous at $x=0$, it cannot be differentiable at $x=0$.
7. Find the sum $a+b$ if the function $f(x) = \begin{cases} e^{3x} & x \le 0 \\ ax+b & x > 0 \end{cases}$ is differentiable for all real numbers.
   Answer: $4$ For the function to be differentiable at $x=0$, it must be continuous and have matching slopes. Continuity requires $e^{3(0)} = a(0)+b$, which simplifies to $1=b$. For the slopes to match, the derivatives of the pieces must be equal at $x=0$. The derivative of $e^{3x}$ is $3e^{3x}$. At $x=0$, this is $3e^0 = 3$. The derivative of $ax+b$ is $a$. Therefore, we must have $a=3$. The sum is $a+b = 3+1=4$.
8. Which statement best describes the function $f(x) = \sqrt[3]{(x-3)^2}$ at the point $x=3$?
   Answer: It is continuous but not differentiable at $x=3$ due to a cusp. The function can be written as $f(x) = (x-3)^{2/3}$. The function is continuous at $x=3$ since $f(3) = (3-3)^{2/3} = 0$. To check for differentiability, we find the derivative: $f'(x) = \frac{2}{3}(x-3)^{-1/3} = \frac{2}{3\sqrt[3]{x-3}}$. As $x$ approaches 3, the denominator approaches 0, so the derivative approaches infinity. Specifically, $\lim_{x \to 3^+} f'(x) = +\infty$ and $\lim_{x \to 3^-} f'(x) = -\infty$. Since the one-sided limits of the derivative are infinite and have opposite signs, the graph has a cusp at $x=3$. Therefore, the function is not differentiable at $x=3$.
9. Let $f$ be a function such that its derivative $f'(1)$ exists. Which of the following statements must be true? I. $f$ is continuous at $x=1$. II. $\lim_{x \to 1} f(x) = f(1)$. III. $\lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$ exists and is finite.
   Answer: I, II, and III The premise is that $f'(1)$ exists, which means the function is differentiable at $x=1$. Statement (III) is the limit definition of the derivative $f'(1)$. Since we are told $f'(1)$ exists, this limit must exist and be a finite number. This is true by definition. A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point. Therefore, statement (I) must be true. Statement (II) is the definition of continuity at $x=1$. Since we concluded that $f$ must be continuous at $x=1$, statement (II) must also be true. Thus, all three statements are necessary consequences of $f$ being differentiable at $x=1$.
10. Which of the following statements about a function $f$ defined on $(-\infty, \infty)$ is guaranteed to be true?
    Answer: If $f$ is not continuous at $x=c$, then $f$ is not differentiable at $x=c$. The fundamental relationship is that differentiability implies continuity. That is, if a function is differentiable at a point, it must be continuous at that point. Statement (C) is the contrapositive of this theorem. The contrapositive of a true statement is always true. If a function is not continuous at a point (e.g., has a jump, hole, or asymptote), its derivative cannot exist at that point. Statement (A) is the converse of the theorem, which is false; for example, $f(x)=|x|$ is continuous at $x=0$ but not differentiable. Statement (B) is false; the existence of a limit doesn't even guarantee continuity (e.g., a removable discontinuity), and continuity does not guarantee differentiability. Statement (D) is false; a function can be defined at a point but still have a jump discontinuity there.
11. Which statement best describes the function $g(x) = (x+2)^{1/3} + 1$ at the point $x=-2$?
    Answer: It is continuous but not differentiable at $x=-2$ due to a vertical tangent. The function $g(x) = \sqrt[3]{x+2} + 1$ is continuous at $x=-2$ because $g(-2) = \sqrt[3]{-2+2} + 1 = 1$. To check differentiability, we find the derivative: $g'(x) = \frac{1}{3}(x+2)^{-2/3} = \frac{1}{3(x+2)^{2/3}}$. As $x$ approaches $-2$ from either the left or the right, the term $(x+2)^{2/3}$ in the denominator approaches 0 through positive values. Thus, $\lim_{x \to -2} g'(x) = +\infty$. When the limit of the derivative is infinite at a point, the function has a vertical tangent there and is not differentiable.
12. If a function $g(x)$ is differentiable at $x=a$, which of the following statements is not necessarily true?
    Answer: The derivative $g'(x)$ is continuous at $x=a$. Differentiability at a point implies continuity at that point. Therefore, statement A must be true. Statement B is the definition of continuity at $x=a$, so it must also be true. Statement C is the definition of the derivative $g'(a)$, so it must be true by the premise. Statement D, however, is not guaranteed. A function can be differentiable at a point, meaning $g'(a)$ exists, without the derivative function $g'(x)$ being continuous at that point. Such a function is called differentiable but not continuously differentiable.
13. Let $g(x) = \begin{cases} f(x) & x \le c \\ m(x-c) + k & x > c \end{cases}$. If $f(x)$ is a differentiable function for all real numbers, what must be the values of $k$ and $m$ so that $g(x)$ is also differentiable at $x=c$?
    Answer: $k = f(c)$ and $m = f'(c)$ For $g(x)$ to be differentiable at $x=c$, it must first be continuous. This requires the limit from the left to equal the limit from the right: $\lim_{x \to c^-} g(x) = f(c)$ and $\lim_{x \to c^+} g(x) = m(c-c)+k = k$. Thus, we must have $k = f(c)$. Next, the derivatives from the left and right must match. The derivative from the left is $f'(c)$. The derivative of the linear function $m(x-c)+k$ is $m$. So, the derivative from the right is $m$. For differentiability, we must have $m = f'(c)$. The line $y=m(x-c)+k$ must be the tangent line to $f(x)$ at $x=c$.
14. Let $f(x)$ be a function that is continuous on the interval $[-2, 2]$. If $f'(0)$ does not exist, which of the following MUST be true?
    Answer: $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ does not exist. The statement '$f'(0)$ does not exist' means precisely that the limit which defines the derivative at $x=0$ does not exist. This limit is $\lim_{h \to 0} \frac{f(0+h)-f(0)}{h}$, which is the same as $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$. A) is incorrect; the function is continuous, so it cannot have a vertical asymptote. C) is incorrect; continuity on $[-2, 2]$ implies $f(0)$ is defined. D) is incorrect; continuity at $x=0$ means $\lim_{x \to 0} f(x) = f(0)$, so the limit must exist.
15. Which of the following functions is continuous for all real numbers but not differentiable at $x=0$?
    Answer: $f(x) = |x| + 1$ A) $f(x) = 1/x$ is not continuous at $x=0$. B) $f(x) = x|x|$ can be written as $x^2$ for $x \ge 0$ and $-x^2$ for $x<0$. Its derivative is $2x$ and $-2x$ respectively. At $x=0$, both derivatives are 0, so it IS differentiable at $x=0$. C) $f(x) = x^{4/3}$. Its derivative is $\frac{4}{3}x^{1/3}$. At $x=0$, the derivative is 0. So it IS differentiable at $x=0$. D) $f(x) = |x|+1$ is the graph of $|x|$ shifted up by 1. It is continuous everywhere, but has a sharp corner at $x=0$ and is therefore not differentiable at $x=0$.
16. Let $g(x) = \sqrt[3]{x^2}$. Which statement best describes the differentiability of $g(x)$ at $x=0$?
    Answer: It is not differentiable at $x=0$ because its graph has a cusp. The function can be written as $g(x) = x^{2/3}$. The function is continuous at $x=0$ since $\lim_{x \to 0} x^{2/3} = 0 = g(0)$. The derivative is $g'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$. The derivative is undefined at $x=0$. Taking the limits of the derivative: $\lim_{x \to 0^+} g'(x) = \frac{2}{3(\text{small pos})} = +\infty$ and $\lim_{x \to 0^-} g'(x) = \frac{2}{3(\text{small neg})} = -\infty$. Since the one-sided limits of the derivative approach opposite infinities, the graph has a cusp at $x=0$, and thus the function is not differentiable there.
17. Let $f(x) = |x-c|$ and $g(x)$ be a function that is differentiable at $x=c$. The product function $h(x) = f(x)g(x)$ is differentiable at $x=c$ if and only if which condition is met?
    Answer: $g(c) = 0$ To check differentiability of $h(x)=|x-c|g(x)$ at $x=c$, we examine the limit of the difference quotient: $h'(c) = \lim_{x \to c} \frac{|x-c|g(x) - |c-c|g(c)}{x-c} = \lim_{x \to c} \frac{|x-c|g(x)}{x-c}$. We check the one-sided limits. From the right (where $x>c$): $\lim_{x \to c^+} \frac{(x-c)g(x)}{x-c} = \lim_{x \to c^+} g(x) = g(c)$. From the left (where $x<c$): $\lim_{x \to c^-} \frac{-(x-c)g(x)}{x-c} = \lim_{x \to c^-} -g(x) = -g(c)$. For the derivative to exist, these limits must be equal: $g(c) = -g(c)$, which implies $2g(c)=0$, or $g(c)=0$.
18. Consider the function $g(x) = \begin{cases} \sin(\pi x) & x \le 1/2 \\ ax+b & x > 1/2 \end{cases}$. If $g(x)$ is differentiable at $x=1/2$, what is the value of $a$?
    Answer: $0$ For $g(x)$ to be differentiable at $x=1/2$, the derivatives of the two pieces must be equal at that point. The derivative of $\sin(\pi x)$ is $\pi\cos(\pi x)$. At $x=1/2$, this derivative is $\pi\cos(\pi/2) = \pi(0) = 0$. The derivative of $ax+b$ is $a$. For the function to be differentiable, we must have $a=0$. We can also find $b$ from the continuity condition: $\sin(\pi/2) = a(1/2)+b$, which gives $1 = (0)(1/2)+b$, so $b=1$. The question only asks for the value of $a$.
19. The function $h(t) = (t^2 - 8t + 16)^{1/3}$ is not differentiable at $t=a$. What is the value of $a$ and why?
    Answer: $a = 4$, because the graph of $h(t)$ has a cusp at $t=4$. First, simplify the expression inside the parentheses: $t^2 - 8t + 16 = (t-4)^2$. So, $h(t) = ((t-4)^2)^{1/3} = (t-4)^{2/3}$. The function is continuous everywhere. The derivative is $h'(t) = \frac{2}{3}(t-4)^{-1/3} = \frac{2}{3\sqrt[3]{t-4}}$. The derivative is undefined at $t=4$. To determine the nature of this point, we check the limits of the derivative: $\lim_{t \to 4^+} h'(t) = +\infty$ and $\lim_{t \to 4^-} h'(t) = -\infty$. Because the one-sided limits of the derivative approach opposite infinities, the graph has a cusp at $t=4$.
20. Let $g(x) = |x^2-4|$ and let $f(x)$ be a function that is differentiable for all real numbers. Let $h(x) = f(x)g(x)$. For which choice of $f(x)$ is $h(x)$ guaranteed to be differentiable at $x=2$?
    Answer: $f(x) = x-2$ The function $g(x) = |x^2-4|$ is not differentiable at $x=2$ because it has a corner there. For the product $h(x)=f(x)g(x)$ to be differentiable at $x=2$, the factor $f(x)$ must 'smooth out' the corner. This happens if $f(2)=0$. We can check this with the limit definition of the derivative: $h'(2) = \lim_{k \to 0} \frac{f(2+k)g(2+k) - f(2)g(2)}{k}$. Since $g(2)=0$, this becomes $\lim_{k \to 0} \frac{f(2+k)|(2+k)^2-4|}{k}$. If $f(2) \ne 0$, this limit will not exist because the term $\frac{|(2+k)^2-4|}{k} = \frac{|k(4+k)|}{k}$ has different left and right limits. However, if $f(2)=0$, then $h'(2)=0$. We must choose the function $f(x)$ for which $f(2)=0$. For choice (D), $f(x)=x-2$, we have $f(2)=2-2=0$. For all other choices, $f(2) \ne 0$.

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