If `f(x) = {{:(x - 3",",x lt 0),(x^(2)-3x + 2",",x ge 0):}"and let" g(x) = f(|x|) + |f(x)|`. Discuss the differentiability of g(x).

Answer

To discuss the differentiability of the function \( g(x) = f(|x|) + |f(x)| \), we first need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} x - 3 & \text{if } x < 0 \\ x^2 - 3x + 2 & \text{if } x \geq 0 \end{cases} \]

### Step 1: Finding \( f(|x|) \)

Since \( |x| \) is always non-negative, we only need to consider the case when \( x \geq 0 \):

\[ f(|x|) = f(x) = x^2 - 3x + 2 \quad \text{for } x \geq 0 \]

For \( x < 0 \), \( |x| = -x \) which is positive, hence:

\[ f(|x|) = f(-x) = -x - 3 \quad \text{for } x < 0 \]

### Step 2: Finding \( |f(x)| \)

Next, we find \( |f(x)| \):

- For \( x < 0 \):
\[ f(x) = x - 3 \implies |f(x)| = |x - 3| = 3 - x \quad \text{(since \( x - 3 < 0 \))} \]

- For \( x \geq 0 \):
\[ f(x) = x^2 - 3x + 2 \]
We need to determine when this expression is positive or negative. The roots of the quadratic are \( x = 1 \) and \( x = 2 \). Thus:
- \( |f(x)| = f(x) = x^2 - 3x + 2 \) for \( x \leq 1 \) and \( x \geq 2 \)
- \( |f(x)| = -(x^2 - 3x + 2) = -x^2 + 3x - 2 \) for \( 1 < x < 2 \)

### Step 3: Constructing \( g(x) \)

Now we can write \( g(x) \):

\[ g(x) = f(|x|) + |f(x)| \]

- For \( x < 0 \):
\[ g(x) = (-x - 3) + (3 - x) = -2x \]

- For \( 0 \leq x < 1 \):
\[ g(x) = (x^2 - 3x + 2) + (x^2 - 3x + 2) = 2(x^2 - 3x + 2) = 2x^2 - 6x + 4 \]

- For \( 1 \leq x < 2 \):
\[ g(x) = (x^2 - 3x + 2) + (-x^2 + 3x - 2) = 0 \]

- For \( x \geq 2 \):
\[ g(x) = (x^2 - 3x + 2) + (x^2 - 3x + 2) = 2(x^2 - 3x + 2) = 2x^2 - 6x + 4 \]

### Step 4: Analyzing Continuity and Differentiability

Now we need to check the continuity and differentiability at the critical points \( x = 0, 1, 2 \).

1. **At \( x = 0 \)**:
- Left limit: \( g(0^-) = 0 \)
- Right limit: \( g(0^+) = 2(0^2 - 3(0) + 2) = 4 \)
- Since \( g(0^-) \neq g(0^+) \), \( g(x) \) is not continuous at \( x = 0 \).

2. **At \( x = 1 \)**:
- Left limit: \( g(1^-) = 2(1^2 - 3(1) + 2) = 0 \)
- Right limit: \( g(1^+) = 0 \)
- Since both limits are equal, \( g(x) \) is continuous at \( x = 1 \).

Checking differentiability:
- Left derivative: \( g'(1^-) = 4(1) - 6 = -2 \)
- Right derivative: \( g'(1^+) = 0 \)
- Since \( g'(1^-) \neq g'(1^+) \), \( g(x) \) is not differentiable at \( x = 1 \).

3. **At \( x = 2 \)**:
- Left limit: \( g(2^-) = 0 \)
- Right limit: \( g(2^+) = 2(2^2 - 3(2) + 2) = 0 \)
- Since both limits are equal, \( g(x) \) is continuous at \( x = 2 \).

Checking differentiability:
- Left derivative: \( g'(2^-) = 0 \)
- Right derivative: \( g'(2^+) = 4(2) - 6 = 2 \)
- Since \( g'(2^-) \neq g'(2^+) \), \( g(x) \) is not differentiable at \( x = 2 \).

### Conclusion

The function \( g(x) \) is not differentiable at \( x = 0, 1, \) and \( 2 \). It is differentiable everywhere else.

---