Let `f:R rarr R, f(x)=x+log_(e)(1+x^(2))`. Then f(x) is what kind of function

Options

A. f is injective
B. f is surjective
C. there is a point on the graph of y= f(x) where tangent is not parallel to any of the chords
D. inverser of f(x) exists.

Explanation

To determine the nature of the function \( f(x) = x + \log_e(1 + x^2) \), we will analyze its monotonicity by finding its first derivative and checking its sign.

### Step 1: Find the first derivative of \( f(x) \)

The function is given by:
\[ f(x) = x + \log_e(1 + x^2) \]

To find the first derivative, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\log_e(1 + x^2)) \]

Using the chain rule for the logarithmic part:
\[ \frac{d}{dx}(\log_e(1 + x^2)) = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2} \]

Thus, the first derivative is:
\[ f'(x) = 1 + \frac{2x}{1 + x^2} \]

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

Now we need to analyze the sign of \( f'(x) \):
\[ f'(x) = 1 + \frac{2x}{1 + x^2} \]

Since \( 1 + x^2 > 0 \) for all \( x \), the term \( \frac{2x}{1 + x^2} \) will determine the sign of \( f'(x) \).

1. If \( x > 0 \): \( \frac{2x}{1 + x^2} > 0 \) implies \( f'(x) > 1 > 0 \) (increasing).
2. If \( x = 0 \): \( f'(0) = 1 + 0 = 1 > 0 \) (increasing).
3. If \( x < 0 \): \( \frac{2x}{1 + x^2} < 0 \), but since \( 1 > |\frac{2x}{1 + x^2}| \) for small negative \( x \), \( f'(x) > 0 \) (increasing).

Thus, \( f'(x) > 0 \) for all \( x \in \mathbb{R} \).

### Step 3: Conclusion about the function \( f(x) \)

Since \( f'(x) > 0 \) for all \( x \), we conclude that the function \( f(x) \) is strictly increasing on \( \mathbb{R} \).

### Final Answer

The function \( f(x) = x + \log_e(1 + x^2) \) is a strictly increasing function.

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