Differentiability and the derivative

Question 1: Definitions of Differentiability (from file image_6652de.png)

The Question: Choose the set of correct options. (Multiple Select Question)

Detailed Solution:

• “If $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists, then $f$ is differentiable at a.”: TRUE. This is the formal definition of differentiability.
• “A function $f$ may be differentiable at a point a, even if it is not continuous at a.”: FALSE. Differentiability requires continuity. If a function is not continuous, it cannot be differentiable.
• “If a function is differentiable at a point a, then it must be continuous at a.”: TRUE. This is the fundamental relationship between the two concepts.
• “There can exist some continuous functions which are not differentiable at some points in the domain.”: TRUE. The classic example is $f(x) = |x|$, which is continuous everywhere but is not differentiable at the sharp corner at $x=0$.

Final Answer: The correct statements are the first, third, and fourth options.

Question 2: Derivative from First Principles (from file image_6652de.png)

The Question: Which of the following options showing step wise solution to check whether a function is differentiable or not are true?

Detailed Solution:

Let’s check each calculation using the limit definition $f’(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

• For $f(x)=c$:

  • $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = 0$.
  • The option says the limit is 1. FALSE.

• For $f(x)=x-c$:

  • $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{((a+h)-c) - (a-c)}{h} = \lim_{h \to 0} \frac{a+h-c-a+c}{h} = \lim_{h \to 0} \frac{h}{h} = 1$.
  • The option says the limit is 1. TRUE.

• For $f(x)=x^2$ (written as $x^0$ in the image, a typo):

  • Assuming $f(x)=x^2$: $\lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} = \lim_{h \to 0} \frac{a^2+2ah+h^2-a^2}{h} = \lim_{h \to 0} \frac{2ah+h^2}{h} = \lim_{h \to 0} (2a+h) = 2a$.
  • The option shows a different calculation that results in 0. FALSE.

• For $f(x)=e^x$:

  • $\lim_{h \to 0} \frac{e^{a+h} - e^a}{h} = \lim_{h \to 0} \frac{e^a e^h - e^a}{h} = \lim_{h \to 0} \frac{e^a(e^h - 1)}{h}$.
  • Using the fundamental limit $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$, the result is $e^a \cdot 1 = e^a$.
  • The option correctly shows these steps and the final result. TRUE.

Final Answer:

• Checking whether $f(x)=x-c$ is differentiable…
• Checking whether $f(x)=e^x$ is differentiable…

Question 3: Differentiability of a Piecewise Function (from file image_664f1c.png)

The Question: Consider the function defined as follows: $f(x) = \begin{cases} |x| & \text{if } x \ge 0 \ \lfloor x \rfloor & \text{if } x < 0 \end{cases}$. Choose the set of correct options.

Detailed Solution:

1. Simplify the function:
   • For $x \ge 0$, $|x| = x$. So, $f(x)=x$ for $x \ge 0$.
   • For $x < 0$, $f(x) = \lfloor x \rfloor$ (the floor function), which is a step function.
2. Analyze Differentiability:
   • At $x=0$: There is a “jump discontinuity”. The right side approaches 0, while the left side approaches -1. Since it’s not continuous, it is not differentiable at x=0.
   • At $x=1.5$: Here, $x > 0$, so we use the rule $f(x)=x$. The graph is a smooth straight line. It is differentiable at x=1.5.
   • At $x=1$: Here, $x > 0$, so we use the rule $f(x)=x$. The graph is a smooth straight line. It is differentiable at x=1.
   • At $x=-1$: This is an integer where the floor function jumps. It is discontinuous here. Therefore, it is not differentiable at x=-1.

Final Answer:

• $f$ is differentiable at $x=1$.
• $f$ is differentiable at $x=1.5$.

Question 4: Differentiability from Graphs (from file image_664edc.png)

The Question: Consider the graphs given below. Choose the set of correct options.

Detailed Solution:

• Figure $f_1$ (V-shape): This is the graph of an absolute value function, likely $y=|x-1|+1$.
  • “f1 is continuous and differentiable at each real number.”: FALSE. It is continuous everywhere, but the sharp corner at $x=1$ means it is not differentiable there.
  • “f1 is not differentiable at 1.”: TRUE. A sharp corner means the slope is not uniquely defined.
• Figure $f_2$ (Curve with a hole):
  • “f2 is not continuous at 0.”: TRUE. There is a hole (a removable discontinuity) at $x=0$.
  • “f2 is differentiable in the interval [1, 2].”: TRUE. In the interval $[1, 2]$, the graph is a smooth, unbroken curve. It is differentiable at every point in this interval.

Final Answer:

• $f_1$ is not differentiable at 1.
• $f_2$ is not continuous at 0.
• $f_2$ is differentiable in the interval [1, 2].

Question 5: Properties of $f(x) = x^{1/3}$ (from file image_664bd7.png)

The Question: The following curve shown in Figure M2W2AQ2 represents the function $f: \mathbb{R} \to \mathbb{R}$, such that $f(x) = x^{1/3}$. Which of the following options is(are) correct?

Detailed Solution:

• "$f$ is continuous on $\mathbb{R}$.": TRUE. The graph can be drawn in one stroke without any breaks or holes.
• "$f$ is differentiable everywhere on $\mathbb{R}$.": FALSE. At $x=0$, the graph becomes perfectly vertical for an instant. A vertical line has an undefined slope. Therefore, the function has a vertical tangent at $x=0$ and is not differentiable there.
• "$f$ is no where differentiable on $\mathbb{R}$.": FALSE. It is differentiable everywhere except at $x=0$.
• "$f$ is not differentiable at 0.": TRUE, due to the vertical tangent.

Final Answer:

• $f$ is continuous on $\mathbb{R}$.
• $f$ is not differentiable at 0.

Question 6: Properties of $f(x)=x|x|$ (from file image_664b74.png)

The Question: Consider the function $f(x) = x|x|$. Which of the following options are correct?

Detailed Solution:

1. Rewrite as a piecewise function:
   • If $x \ge 0$, $|x|=x$, so $f(x) = x(x) = x^2$.
   • If $x < 0$, $|x|=-x$, so $f(x) = x(-x) = -x^2$. $f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \ -x^2 & \text{if } x < 0 \end{cases}$
2. Check Continuity: The two pieces are parabolas, which are continuous. We only need to check the breakpoint at $x=0$.
   • Right limit: $\lim_{x \to 0^+} x^2 = 0$.
   • Left limit: $\lim_{x \to 0^-} -x^2 = 0$.
   • Function value: $f(0)=0^2=0$.
   • Since all three are equal, the function is continuous on $\mathbb{R}$.
3. Check Differentiability at $x=0$: We check if the derivatives from the left and right are equal.
   • Derivative from the right (for $x>0$): The derivative of $x^2$ is $2x$. At $x=0$, the slope is $2(0)=0$.
   • Derivative from the left (for $x<0$): The derivative of $-x^2$ is $-2x$. At $x=0$, the slope is $-2(0)=0$.
   • Since the left-derivative and right-derivative are both 0, the function is differentiable at $x=0$.
4. Conclusion: The function is continuous and differentiable on all of $\mathbb{R}$.

Final Answer: $f$ is both continuous and differentiable on $\mathbb{R}$.

Question 7: Advanced Differentiability Concepts (from file image_664b74.png)

The Question: Choose the set of correct options.

Detailed Solution:

• “There exists a function $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is not differentiable exactly on the set of natural numbers”: TRUE. This is possible. The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere. More simply, one can construct a function by summing “sawtooth” functions centered at each natural number.
• “Inverse of a differentiable function is differentiable.”: FALSE. A counterexample is $f(x)=x^3$. It is differentiable everywhere. Its inverse is $f^{-1}(x) = x^{1/3}$, which is not differentiable at $x=0$. This happens when the original function has a point with a horizontal tangent ($f’(x)=0$).
• “Let $f(x) = x^2$ for x rational and $f(x)=0$ for x irrational. Then $f$ is differentiable at $x=0$.”: TRUE. This is a famous example. Using the limit definition of the derivative at $a=0$, we get $\lim_{h \to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$. If $h$ is rational, this is $\frac{h^2}{h}=h$. If $h$ is irrational, this is $\frac{0}{h}=0$. In both cases, as $h \to 0$, the limit is 0. So the derivative exists and is 0.

Final Answer:

• There exists a function $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is not differentiable exactly on the set of natural numbers
• Let $f(x) = x^2$ for x rational and $f(x)=0$ for x irrational. Then $f$ is differentiable at $x=0$.

Question 8: Properties of $f(x)=|\sin x|$ (from file image_664b74.png)

The Question: Consider the function $f(x)=|\sin x|$. Then $f$ is…

Detailed Solution:

• Periodicity: The function $\sin x$ has a period of $2\pi$. The absolute value function $|\sin x|$ reflects the negative parts of the sine wave upwards. The resulting shape now repeats every $\pi$ units. The function is periodic with period $\pi$.
• Continuity: The function $\sin x$ is continuous everywhere. The absolute value function is also continuous everywhere. The composition of two continuous functions is continuous. So, $f(x)=|\sin x|$ is everywhere continuous.
• Differentiability: The graph of $|\sin x|$ has sharp corners wherever the original $\sin x$ graph crossed the x-axis (except where it was already tangent). This happens when $\sin x = 0$, which is at $x = n\pi$ for every integer $n$ (…, $-\pi, 0, \pi, 2\pi$, …). At these sharp points, the function is not differentiable.

Final Answer:

• periodic with period $\pi$.
• everywhere continuous.
• everywhere continuous and not differentiable at $n\pi$, where $n \in \mathbb{Z}$.

Question 9: Derivative from a Graph (from file image_664b15.png)

The Question: Consider Figure M2W2AQ3, which represents some function $f$, to choose the correct options from the following.

Detailed Solution: The graph shows a smooth, wave-like function (like a sine or cosine wave).

• “There are some points in the interval $[-5, 5]$ at which the derivative of $f$ is 0.”: TRUE. The derivative is the slope of the tangent line. The slope is 0 at every peak (local maximum) and valley (local minimum). The graph clearly shows several peaks and valleys in the interval $[-5, 5]$.
• “There are some points in the interval $[-5, 5]$ at which $f$ is not differentiable.”: FALSE. The graph shown is perfectly smooth and continuous, with no corners, cusps, or vertical tangents. It appears to be differentiable everywhere.
• "$f$ is not differentiable at 0.": FALSE. The graph is smooth at $x=0$.
• “There are at least 2 points where the derivative of $f$ are non-zero but the same.”: TRUE. For example, the slope at $x=0$ is positive and steep. Due to the wave-like nature, there will be another point (e.g., near $x=2\pi \approx 6.28$, or more relevantly, near $x=-2\pi \approx -6.28$, and also around $x=0$ but on the other side of the peak) where the slope is the same. Even more simply, take a point on an upslope, say at $x=-4$. The slope is positive. There’s another upslope around $x=2$ that will have a point with the exact same positive slope.

Final Answer:

• There are some points in the interval $[-5, 5]$ at which the derivative of $f$ is 0.
• There are at least 2 points where the derivative of $f$ are non-zero but the same.

Question 10: Differentiable Piecewise Function (from file image_664b15.png)

The Question: Define a function $f$ as follows: $f(x) = \begin{cases} kx^2 + l & \text{if } x \le 1 \ lx^2 + kx + m & \text{if } x > 1 \end{cases}$. If $f$ is continuous and differentiable at $x=1$ then the value of $k-2l+m$ is __________.

Detailed Solution: The function must be continuous and differentiable at the breakpoint $x=1$.

1. Continuity Condition: The values of the two pieces must be equal at $x=1$.

   • From the top piece: $f(1) = k(1)^2 + l = k+l$.
   • From the bottom piece (as a limit): $\lim_{x \to 1^+} (lx^2+kx+m) = l(1)^2+k(1)+m = l+k+m$.
   • Set them equal: $k+l = l+k+m$.
   • This simplifies to $m=0$.

2. Differentiability Condition: The derivatives of the two pieces must be equal at $x=1$.

   • Find the derivative of the top piece: $\frac{d}{dx}(kx^2+l) = 2kx$. At $x=1$, this is $2k(1) = 2k$.
   • Find the derivative of the bottom piece: $\frac{d}{dx}(lx^2+kx+m) = 2lx+k$. At $x=1$, this is $2l(1)+k = 2l+k$.
   • Set them equal: $2k = 2l+k$.
   • This simplifies to $k = 2l$.

3. Calculate the final expression:

   • We need to find the value of $k-2l+m$.
   • Substitute the relationships we found: $k=2l$ and $m=0$.
   • $(2l) - 2l + 0 = 0$.

Final Answer: The value of $k-2l+m$ is 0.