If `g(x)=int_0^x2|t|dt ,t h e n`
(a) `g(x)=x|x|` (b)`g(x)` is monotonic
(c)`g(x)` is differentiable at `x=0` (d)`g^(prime)(x)` is differentiable at `x=0`

Options

A. `g(x)=x|x|`
B. g(x) is monotonic
C. g(x) is differentiable at x = 0
D. g'(x) is differentiable at x = 0

Answer

To solve the problem, we will evaluate the function \( g(x) \) defined by the integral \( g(x) = \int_0^x 2|t| \, dt \) and analyze its properties.

### Step 1: Evaluate the integral for \( g(x) \)

The function \( |t| \) behaves differently based on the sign of \( t \). Therefore, we will consider two cases:

1. **Case 1**: When \( x \geq 0 \)
\[ g(x) = \int_0^x 2t \, dt \]
Evaluating this integral:
\[ g(x) = 2 \left[ \frac{t^2}{2} \right]_0^x = [t^2]_0^x = x^2 \]

2. **Case 2**: When \( x < 0 \)
\[ g(x) = \int_0^x 2(-t) \, dt = -\int_0^x 2t \, dt = -\left[ t^2 \right]_0^x = -x^2 \]

Combining both cases, we can express \( g(x) \) as:
\[ g(x) = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \]

### Step 2: Express \( g(x) \) in terms of \( |x| \)

We can rewrite \( g(x) \) using the absolute value:
\[ g(x) = x|x| \]
This holds true since:
- For \( x \geq 0 \), \( g(x) = x^2 = x|x| \)
- For \( x < 0 \), \( g(x) = -x^2 = x|x| \)

### Step 3: Check if \( g(x) \) is monotonic

To determine if \( g(x) \) is monotonic, we will find the derivative \( g'(x) \):
1. For \( x \geq 0 \):
\[ g'(x) = 2x \]
2. For \( x < 0 \):
\[ g'(x) = -2x \]

Since \( g'(x) \) is positive for \( x > 0 \) and negative for \( x < 0 \), \( g(x) \) is increasing for \( x \geq 0 \) and decreasing for \( x < 0 \). Thus, \( g(x) \) is a monotonic function.

### Step 4: Check differentiability at \( x = 0 \)

To check if \( g(x) \) is differentiable at \( x = 0 \), we will find the left-hand and right-hand derivatives:
- Right-hand derivative at \( x = 0 \):
\[ g'_+(0) = \lim_{h \to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0 \]

- Left-hand derivative at \( x = 0 \):
\[ g'_-(0) = \lim_{h \to 0^-} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} -h = 0 \]

Since both derivatives are equal, \( g(x) \) is differentiable at \( x = 0 \).

### Step 5: Check differentiability of \( g'(x) \) at \( x = 0 \)

Now we will check if \( g'(x) \) is differentiable at \( x = 0 \):
1. For \( x \geq 0 \):
\[ g'(x) = 2x \quad \Rightarrow \quad g''(x) = 2 \]
2. For \( x < 0 \):
\[ g'(x) = -2x \quad \Rightarrow \quad g''(x) = -2 \]

The right-hand derivative at \( x = 0 \) is:
\[ g''_+(0) = 2 \]
The left-hand derivative at \( x = 0 \) is:
\[ g''_-(0) = -2 \]

Since the left-hand and right-hand derivatives are not equal, \( g'(x) \) is not differentiable at \( x = 0 \).

### Conclusion

Based on the analysis:
- (a) \( g(x) = x|x| \) is **true**.
- (b) \( g(x) \) is monotonic is **true**.
- (c) \( g(x) \) is differentiable at \( x = 0 \) is **true**.
- (d) \( g'(x) \) is differentiable at \( x = 0 \) is **false**.