Let f : R → R be a function defined as f(x) = 3(1 - |x|/2) if |x| ≤ 2 and 0 if |x| 2. Let g : R →

Q: Wait, why is $n=0$? Shouldn't there be a discontinuity when we shift the functions?

Great question! Remember that $f(x)$ is continuous everywhere. Shifting a continuous function horizontally (like $f(x+2)$ or $f(x-2)$) simply moves the graph; it does not introduce any breaks or jumps. Since the difference of two continuous functions is also continuous, $g(x)$ remains continuous for all $x \in \mathbb{R}$. Thus, $n=0$.

Q: Why do we check $x=0$ if it's a corner of both $f(x+2)$ and $f(x-2)$? Doesn't that automatically mean it's non-differentiable?

That is a classic trap! While $x=0$ is a sharp corner for both $f(x+2)$ and $f(x-2)$, we must look at the *combined* function $g(x)$. It is possible for the 'sharpness' of the two components to cancel each other out. By calculating the LHD and RHD at $x=0$, we find that the slopes actually match, making the function smooth at that specific point.

Q: How do I handle the absolute value when calculating derivatives?

The best way is to define the function piecewise in the neighborhood of the point in question. For example, near $x=0$, $x+2$ is positive, so $|x+2| = x+2$, and $x-2$ is negative, so $|x-2| = -(x-2)$. Once you remove the absolute value bars by substituting these expressions, you get simple linear equations that are easy to differentiate.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined as $f(x) = \begin{cases} 3 (1 - \frac{|x|}{2}) & 	ext{if } |x| \le 2 \ 0 & 	ext{if } |x| > 2 \end{cases}$. Let $g : \mathbb{R} 	o \mathbb{R}$ be given by $g(x) = f(x + 2) - f(x - 2)$. If $n$ and $m$ denote the number of points in $\mathbb{R}$ where $g$ is not continuous and not differentiable, respectively, then $n + m$ is equal to ____.

Sigma Percentile · Accepted Answer

### The Geometry of the Pulse

Imagine you are standing on a vast, flat plain, and suddenly, a perfectly symmetric triangular mountain rises from the ground. This is the visual soul of our function:

$$f(x) = 3\left(1 - \frac{|x|}{2}ight) \quad 	ext{for } |x| \le 2, 	ext{ and } 0 	ext{ otherwise.}$$

It peaks at $y=3$ when $x=0$ and gently slopes down to touch the x-axis at $x=-2$ and $x=2$.

It is a beautiful, continuous shape, but it hides a secret: those sharp peaks and valleys at $x=-2, 0, 2$ are where the derivative—the measure of the slope—refuses to exist. In the world of calculus, these are our **points of non-differentiability**.

### The Dance of the Shifts

Now, we introduce a new player: $g(x) = f(x+2) - f(x-2)$. Think of this as a choreographed dance.

$f(x+2)$ is our original triangular pulse shifted $2$ units to the left, while $f(x-2)$ is the same pulse shifted $2$ units to the right. We are subtracting one from the other.

Because $f(x)$ is continuous everywhere, shifting it does not break it. Since the difference of two continuous functions is always continuous, we can confidently state that $g(x)$ has no points of discontinuity. Our first variable, $n$, is firmly $0$.

### The Differentiability Hunt

This is where the plot thickens. We know that $g(x)$ can only be non-differentiable where its components, $f(x+2)$ or $f(x-2)$, are non-differentiable.

Let's track the "sharp corners." The original corners were at $\{-2, 0, 2\}$.

Shifting $f(x)$ left by $2$ units moves these corners to $\{-4, -2, 0\}$. Shifting it right by $2$ units moves them to $\{0, 2, 4\}$. Our "suspect list" for non-differentiability is the union of these sets: $\{-4, -2, 0, 2, 4\}$.

### The $x=0$ Revelation

Many students would stop here and count five points. But in JEE Advanced, we must be precise. Let's look at $x=0$.

Is it truly a sharp corner for $g(x)$? To find out, we calculate the slopes on either side.

For $x$ just to the left of $0$ (in the interval $(-2, 0)$), $g(x)$ behaves like $f(x+2)$, which simplifies to a linear expression with slope $-1.5$. For $x$ just to the right of $0$ (in the interval $(0, 2)$), $g(x)$ behaves like $-f(x-2)$, which also simplifies to a linear expression with slope $-1.5$.

Because the Left-Hand Derivative (LHD) equals the Right-Hand Derivative (RHD), the function $g(x)$ is actually smooth at $x=0$. The sharp corners of the individual pulses have perfectly cancelled each other out, creating a straight line passing through the origin.

### The Final Tally

With $x=0$ cleared of suspicion, we are left with the remaining points: $\{-4, -2, 2, 4\}$. These are the true locations where the slope changes abruptly.

Thus, the number of points of non-differentiability, $m$, is $4$.

Since $n=0$, our final answer is:

$$n + m = 0 + 4 = \mathbf{4}$$

You have successfully navigated the trap, visualized the geometry, and arrived at the truth. Keep this intuition close; it is the key to mastering the most challenging problems in calculus.

Enter Numerical Value:

Visualized Solution

Understanding $f (x)$

Continuity of $f (x)$

Continuity of $g (x)$

Non-differentiability of $f (x)$

Points for $f (x + 2)$

Points for $f (x - 2)$

Potential Points for $g (x)$

Visualizing $g (x)$

Checking $x = 0$ (LHD)

Checking $x = 0$ (RHD)

Differentiability at $x = 0$

Final Calculation

The Sigma Insight: Relationship Between Continuity and Differentiability

The Geometry of the Pulse

The Dance of the Shifts

The Differentiability Hunt

The $x = 0$ Revelation

The Final Tally

Similar Questions

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Let $f (x) = ⎩ ⎨ ⎧ 3 x, min {1 + x + [x], x + 2 [x]}, 5, x < 0 0 \leq x \leq 2 x > 2,$ where [.] denotes greatest integer function. If $α$ and $β$ are the number of points, where f is not continuous and is not differentiable, respectively, then $α + β$ equals

Enter Numerical Value:

Visualized Solution

Understanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Continuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Continuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Non-differentiability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x+2)

Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x−2)

Potential Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 (LHD)

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 (RHD)

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0