`f: (0,oo) to (-pi/2,pi/2)" be defined as, "f(x)=tan^(-1) (log_(e)x)`.
The graph of y = f (x) is best represented by

Options

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Answer

To analyze the function \( f(x) = \tan^{-1}(\log_e x) \) and determine the behavior of its graph, we will follow these steps:

### Step 1: Determine the limit of \( f(x) \) as \( x \) approaches 0 from the right.

We start by evaluating \( f(x) \) as \( x \) approaches 0 from the right:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \tan^{-1}(\log_e x) \]

As \( x \) approaches 0, \( \log_e x \) approaches \( -\infty \). Therefore:

\[ \lim_{x \to 0^+} f(x) = \tan^{-1}(-\infty) = -\frac{\pi}{2} \]

### Step 2: Determine the limit of \( f(x) \) as \( x \) approaches infinity.

Next, we evaluate \( f(x) \) as \( x \) approaches infinity:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \tan^{-1}(\log_e x) \]

As \( x \) approaches infinity, \( \log_e x \) approaches \( +\infty \). Therefore:

\[ \lim_{x \to \infty} f(x) = \tan^{-1}(\infty) = \frac{\pi}{2} \]

### Step 3: Analyze the behavior of \( f(x) \) in the interval \( (0, \infty) \).

Since \( f(x) \) is continuous in the interval \( (0, \infty) \) and we have established the limits at both ends:

- As \( x \to 0^+ \), \( f(x) \to -\frac{\pi}{2} \)
- As \( x \to \infty \), \( f(x) \to \frac{\pi}{2} \)

### Step 4: Conclusion about the graph of \( f(x) \).

The function \( f(x) \) is increasing because the logarithm function is increasing and the arctangent function is also increasing. Therefore, the graph of \( f(x) \) will start from \( -\frac{\pi}{2} \) at \( x = 0 \) and approach \( \frac{\pi}{2} \) as \( x \) goes to infinity.

### Final Answer:

The graph of \( y = f(x) \) is best represented by a curve that starts from \( -\frac{\pi}{2} \) and approaches \( \frac{\pi}{2} \) as \( x \) increases.

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