Let `g(x)=int_0^(e^x) (f\'(t)dt)/(1+t^2)`. Which of the following is true?

Options

A. `g'(x)` is positive on`(-oo,0)` and negative on `(0,oo)`
B. `g'(x)` is negative on`(-oo,0)` and positive on `(0,oo)`
C. `g'(x)` change sign on both `(-oo,0) and (0,oo)`
D. `g'(x)` does not change sign on `(-oo,oo)`.

Answer

To solve the problem, we need to analyze the function \( g(x) \) given by:

\[ g(x) = \int_0^{e^x} \frac{f'(t)}{1 + t^2} \, dt \]

We want to find the derivative \( g'(x) \) and determine its sign over the intervals \( (-\infty, 0) \) and \( (0, \infty) \).

### Step 1: Differentiate \( g(x) \)

To find \( g'(x) \), we can use the Fundamental Theorem of Calculus and the chain rule. The derivative of an integral with variable limits is given by:

\[ g'(x) = \frac{d}{dx} \left( \int_0^{e^x} \frac{f'(t)}{1 + t^2} \, dt \right) \]

Using the chain rule, we get:

\[ g'(x) = \frac{f'(e^x)}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \]

Since \( \frac{d}{dx}(e^x) = e^x \), we can simplify this to:

\[ g'(x) = \frac{f'(e^x) \cdot e^x}{1 + e^{2x}} \]

### Step 2: Analyze the sign of \( g'(x) \)

Now we need to analyze the sign of \( g'(x) \):

\[ g'(x) = \frac{f'(e^x) \cdot e^x}{1 + e^{2x}} \]

- The term \( e^x \) is always positive for all \( x \).
- The denominator \( 1 + e^{2x} \) is also always positive for all \( x \).

Thus, the sign of \( g'(x) \) depends solely on the sign of \( f'(e^x) \).

### Step 3: Determine the intervals for \( f'(e^x) \)

1. **For \( x < 0 \)**:
- As \( x \) approaches \( -\infty \), \( e^x \) approaches \( 0 \) (from the left).
- Hence, \( e^x \) will take values in the interval \( (0, 1) \).
- If \( f'(t) \) is negative in this interval, then \( g'(x) < 0 \).

2. **For \( x > 0 \)**:
- As \( x \) approaches \( +\infty \), \( e^x \) approaches \( +\infty \).
- Hence, \( e^x \) will take values greater than \( 1 \).
- If \( f'(t) \) is positive in this interval, then \( g'(x) > 0 \).

### Conclusion

- Therefore, if \( f'(t) < 0 \) for \( t \in (0, 1) \) and \( f'(t) > 0 \) for \( t > 1 \), we conclude that:
- \( g'(x) < 0 \) for \( x < 0 \)
- \( g'(x) > 0 \) for \( x > 0 \)

Thus, the correct option is:

**g'(x) is negative on \((-∞, 0)\) and positive on \((0, ∞)\)**.