Question:
Let \(f : \mathbb{R} \to \mathbb{R}\) be a function defined as
\(f(x) = \begin{cases} 3\left(1 - \frac{|x|}{2}\right) & \text{if } |x| \le 2 \\ 0 & \text{if } |x| \gt 2 \end{cases}\)
Let \(g : \mathbb{R} \to \mathbb{R}\) 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 ______________.
Let \(f : \mathbb{R} \to \mathbb{R}\) be a function defined as
\(f(x) = \begin{cases} 3\left(1 - \frac{|x|}{2}\right) & \text{if } |x| \le 2 \\ 0 & \text{if } |x| \gt 2 \end{cases}\)
Let \(g : \mathbb{R} \to \mathbb{R}\) 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 ______________.
\(f(x) = \begin{cases} 3\left(1 - \frac{|x|}{2}\right) & \text{if } |x| \le 2 \\ 0 & \text{if } |x| \gt 2 \end{cases}\)
Let \(g : \mathbb{R} \to \mathbb{R}\) 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 ______________.
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\(f\) is a continuous tent on \([-2, 2]\); write \(g\) piecewise and check slopes at the junctions \(-4, -2, 0, 2, 4\).
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
\(f\) is continuous everywhere (it equals 0 at \(x = \pm 2\)), so \(g\) is continuous everywhere and \(n = 0\).
Piecewise form of \(g\):
\(x \in [-4, -2]\): \(g = \frac{3(x + 4)}{2}\), slope \(\frac32\).
\(x \in [-2, 0]\): \(g = -\frac{3x}{2}\), slope \(-\frac32\).
\(x \in [0, 2]\): \(g = -\frac{3x}{2}\), slope \(-\frac32\).
\(x \in [2, 4]\): \(g = -\frac{3(4 - x)}{2}\), slope \(\frac32\).
Outside \([-4, 4]\): \(g = 0\), slope 0.
Slope jumps: at \(x = -4\) (0 to \(\frac32\)), \(x = -2\) (\(\frac32\) to \(-\frac32\)), \(x = 2\) (\(-\frac32\) to \(\frac32\)), \(x = 4\) (\(\frac32\) to 0): not differentiable. At \(x = 0\) both sides have slope \(-\frac32\), so \(g\) is differentiable there.
So \(m = 4\) and \(n + m = 4\).
Answer: 4
Piecewise form of \(g\):
\(x \in [-4, -2]\): \(g = \frac{3(x + 4)}{2}\), slope \(\frac32\).
\(x \in [-2, 0]\): \(g = -\frac{3x}{2}\), slope \(-\frac32\).
\(x \in [0, 2]\): \(g = -\frac{3x}{2}\), slope \(-\frac32\).
\(x \in [2, 4]\): \(g = -\frac{3(4 - x)}{2}\), slope \(\frac32\).
Outside \([-4, 4]\): \(g = 0\), slope 0.
Slope jumps: at \(x = -4\) (0 to \(\frac32\)), \(x = -2\) (\(\frac32\) to \(-\frac32\)), \(x = 2\) (\(-\frac32\) to \(\frac32\)), \(x = 4\) (\(\frac32\) to 0): not differentiable. At \(x = 0\) both sides have slope \(-\frac32\), so \(g\) is differentiable there.
So \(m = 4\) and \(n + m = 4\).
Answer: 4
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