If `f(x)={{:((x^(2)-1)/(x^(2)+1),,",",0ltxle2),((1)/(4)(x^(3)-x^(2)),,",",2ltxle3),((9)/(4)(abs(x-4)+abs(2-x)),,",",3ltxlt4):}` then

Options

A. f(x) is differentiable at x = 2 & x = 3
B. f (x) is non–differentiable at x = 2 & x = 3
C. f(x) is differentiable at x = 3 but not at x = 2
D. f(x) is differentiable at x = 2 but not at x = 3.

Answer

To solve the problem of determining the differentiability of the piecewise function defined as:

\[ f(x) = \begin{cases} \frac{x^2 - 1}{x^2 + 1} & \text{for } 0 < x \leq 2 \\ \frac{1}{4}(x^3 - x^2) & \text{for } 2 < x \leq 3 \\ \frac{9}{4}(|x - 4| + |2 - x|) & \text{for } 3 < x < 4 \end{cases} \]

we will follow these steps:

### Step 1: Find the derivative for each interval

1. **For \(0 < x \leq 2\)**:
\[ f(x) = \frac{x^2 - 1}{x^2 + 1} \]
Using the quotient rule:
\[ f'(x) = \frac{(x^2 + 1)(2x) - (x^2 - 1)(2x)}{(x^2 + 1)^2} \]
Simplifying:
\[ f'(x) = \frac{2x(x^2 + 1 - (x^2 - 1))}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2} \]

2. **For \(2 < x \leq 3\)**:
\[ f(x) = \frac{1}{4}(x^3 - x^2) \]
Differentiating:
\[ f'(x) = \frac{1}{4}(3x^2 - 2x) = \frac{x(3x - 2)}{4} \]

3. **For \(3 < x < 4\)**:
\[ f(x) = \frac{9}{4}(|x - 4| + |2 - x|) \]
In this interval, \(x - 4 < 0\) and \(2 - x < 0\), so:
\[ f(x) = \frac{9}{4}(-x + 4 - x + 2) = \frac{9}{4}(6 - 2x) = \frac{27}{2} - \frac{9}{2}x \]
Differentiating:
\[ f'(x) = -\frac{9}{2} \]

### Step 2: Check differentiability at the boundaries \(x = 2\) and \(x = 3\)

1. **At \(x = 2\)**:
- Right-hand derivative:
\[ f'_{R}(2) = \lim_{x \to 2^+} f'(x) = \lim_{x \to 2^+} \frac{x(3x - 2)}{4} = \frac{2(3 \cdot 2 - 2)}{4} = \frac{2(6 - 2)}{4} = \frac{8}{4} = 2 \]
- Left-hand derivative:
\[ f'_{L}(2) = \lim_{x \to 2^-} f'(x) = \lim_{x \to 2^-} \frac{4x}{(x^2 + 1)^2} = \frac{4 \cdot 2}{(2^2 + 1)^2} = \frac{8}{25} \]
Since \(f'_{R}(2) \neq f'_{L}(2)\), \(f\) is not differentiable at \(x = 2\).

2. **At \(x = 3\)**:
- Right-hand derivative:
\[ f'_{R}(3) = \lim_{x \to 3^+} f'(x) = -\frac{9}{2} \]
- Left-hand derivative:
\[ f'_{L}(3) = \lim_{x \to 3^-} f'(x) = \lim_{x \to 3^-} \frac{x(3x - 2)}{4} = \frac{3(3 \cdot 3 - 2)}{4} = \frac{3(9 - 2)}{4} = \frac{21}{4} \]
Since \(f'_{R}(3) \neq f'_{L}(3)\), \(f\) is not differentiable at \(x = 3\).

### Conclusion
The function \(f(x)\) is not differentiable at both \(x = 2\) and \(x = 3\).