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Step 1: Understand the Definition of f(x)
The function f is given by
f(x) =
\begin{cases}
3\!\Bigl(1 - \frac{|x|}{2}\Bigr), & \text{if } |x|\le 2,\\
0, & \text{if } |x|>2.
\end{cases}
This is a "tent"-shaped function centered at x = 0, taking its maximum value 3 at x = 0 and going to 0 at x = ±2.
Step 2: Rewrite f(x) in Piecewise Linear Form
For more precise interval-by-interval analysis, note that:
For −2 ≤ x < 0, we have |x| = −x, so
f(x) = 3\Bigl(1 - \frac{-x}{2}\Bigr) = 3\Bigl(1 + \frac{x}{2}\Bigr) = \frac{3}{2}(x + 2).
For 0 ≤ x ≤ 2, we have |x| = x, so
f(x) = 3\Bigl(1 - \frac{x}{2}\Bigr) = \frac{3}{2}(2 - x).
For |x| > 2, f(x) = 0.
So, in expanded piecewise form:
f(x) =
\begin{cases}
0, & x < -2, \\
\frac{3}{2}(x + 2), & -2 \le x < 0, \\
\frac{3}{2}(2 - x), & 0 \le x \le 2, \\
0, & x > 2.
\end{cases}
Step 3: Define g(x) = f(x+2) − f(x−2)
To study g(x), we will substitute x+2 and x−2 into the piecewise form of f, then take their difference:
3.1. Find f(x+2)
Let y = x + 2. Then f(y) = f(x+2) is nonzero only if |x+2| ≤ 2, i.e., −2 ≤ x+2 ≤ 2, which simplifies to −4 ≤ x ≤ 0. Within that interval, we further split at y = 0 → x = −2.
Hence:
If x < −4, then x+2 < −2, so f(x+2) = 0.
For −4 ≤ x < −2, we are in the region −2 ≤ y < 0 for f(y), so
f(x+2) = \frac{3}{2}\bigl((x+2)+2\bigr) = \frac{3}{2}(x+4).
For −2 ≤ x ≤ 0, we are in the region 0 ≤ y ≤ 2 for f(y), so
f(x+2) = \frac{3}{2}\bigl(2 - (x+2)\bigr) = \frac{3}{2}(0 - x) = -\frac{3}{2}x.
If x > 0, then x+2 > 2, so f(x+2) = 0.
3.2. Find f(x−2)
Let z = x − 2. Then |z| ≤ 2 means −2 ≤ x−2 ≤ 2, or 0 ≤ x ≤ 4. We split accordingly:
If x < 0, then x−2 < −2, so f(x−2) = 0.
For 0 ≤ x < 2, we have −2 ≤ z < 0 for f(z), so
f(x-2) = \frac{3}{2}\bigl((x-2)+2\bigr) = \frac{3}{2}x.
For 2 ≤ x ≤ 4, we have 0 ≤ z ≤ 2 for f(z), so
f(x-2) = \frac{3}{2}\bigl(2 - (x-2)\bigr) = \frac{3}{2}(4 - x).
If x > 4, then x−2 > 2, so f(x−2) = 0.
Step 4: Construct g(x) = f(x+2) − f(x−2)
Combining the above piecewise expressions carefully, we examine the intervals determined by the critical points x = −4, −2, 0, 2, 4:
x < −4:
f(x+2) = 0, f(x−2) = 0, so g(x) = 0.
−4 ≤ x < −2:
f(x+2) = \frac{3}{2}(x+4) , f(x−2) = 0,
so g(x) = \frac{3}{2}(x+4) .
−2 ≤ x < 0:
f(x+2) = -\frac{3}{2}x , f(x−2) = 0,
so g(x) = -\frac{3}{2}x .
0 ≤ x < 2:
f(x+2) = 0, f(x−2) = \frac{3}{2}x ,
so g(x) = 0 - \frac{3}{2}x = -\frac{3}{2}x .
2 ≤ x ≤ 4:
f(x+2) = 0, f(x−2) = \frac{3}{2}(4 - x) ,
so g(x) = 0 - \frac{3}{2}(4 - x) = -\frac{3}{2}(4 - x) = \frac{3}{2}x - 6.
x > 4:
f(x+2) = 0, f(x−2) = 0, so g(x) = 0.
Step 5: Check Continuity of g(x)
Potential discontinuities occur only at boundary points x = −4, −2, 0, 2, and 4. We check left-hand and right-hand limits (and function values) at these points:
At x = −4, both sides yield g = 0 or \frac{3}{2}(x+4) at x = −4, which is 0. So continuous.
At x = −2, from the left we get \frac{3}{2}(x+4) , at x = −2 → \frac{3}{2}(2) = 3;
from the right, we get -\frac{3}{2}x at x = −2 → 3. Same value. So continuous.
At x = 0, from the left: -\frac{3}{2}\cdot0 = 0; from the right: -\frac{3}{2}\cdot0 = 0. So continuous.
At x = 2, from the left: -\frac{3}{2}\cdot2 = -3; from the right: \frac{3}{2}\cdot2 - 6 = 3 - 6 = -3. So continuous.
At x = 4, from the left: \frac{3}{2}\cdot4 - 6 = 6 - 6 = 0, and from the right it is also 0. So continuous.
Hence, g(x) is continuous everywhere (n = 0).
Step 6: Check Differentiability of g(x)
Non-differentiable points often occur at these same boundary points where the piecewise definition changes slope. We compute the left and right derivatives at x = −4, −2, 0, 2, and 4:
x = −4:
Left of −4, g(x) = 0 → derivative = 0.
Right of −4, g(x) = \frac{3}{2}(x+4) → slope = \frac{3}{2} .
Different slopes → non-differentiable at x = −4.
x = −2:
Left slope for \frac{3}{2}(x+4) is \frac{3}{2} .
Right slope for -\frac{3}{2}x is -\frac{3}{2} .
Not equal → non-differentiable at x = −2.
x = 0:
Slope changes from -\frac{3}{2} (for −2 ≤ x < 0) to -\frac{3}{2} (for 0 ≤ x < 2).
Here both sides have the same slope − \frac{3}{2} → differentiable at x = 0.
x = 2:
Left slope is − \frac{3}{2} (for 0 ≤ x < 2).
Right slope for \frac{3}{2}(x) - 6 is \frac{3}{2} .
Not equal → non-differentiable at x = 2.
x = 4:
Left slope is slope of \frac{3}{2}x - 6 , which is \frac{3}{2} .
Right slope for x > 4 is 0.
Not equal → non-differentiable at x = 4.
We see four non-differentiable points at x = −4, −2, 2, and 4. Thus m = 4.
Step 7: Final Answer
We have found:
n (discontinuities of g) = 0.
m (non-differentiable points of g) = 4.
Therefore,
n + m = 0 + 4 = 4.
