Let f(x) = x log x ` +1` then the set {x : `f(x) gt0`} is equal to

Options

A. `(1,oo)`
B. `(1//e, oo)`
C. `[e, oo)`
D. `(0, 1) uu (1,oo)`

Answer

To solve the problem, we need to find the set of values \( x \) for which \( f(x) = x \log x + 1 > 0 \).

### Step 1: Set up the inequality
We start with the inequality:
\[ f(x) = x \log x + 1 > 0 \]
This can be rearranged to:
\[ x \log x > -1 \]

### Step 2: Analyze the function
Next, we need to analyze the function \( f(x) = x \log x + 1 \). We will first find the derivative to determine the behavior of the function.

### Step 3: Differentiate \( f(x) \)
The derivative of \( f(x) \) is:
\[ f'(x) = \frac{d}{dx}(x \log x) + \frac{d}{dx}(1) = \log x + 1 \]
This is derived using the product rule where \( u = x \) and \( v = \log x \).

### Step 4: Find critical points
To find the critical points, we set the derivative equal to zero:
\[ \log x + 1 = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e} \]

### Step 5: Determine the intervals
We will analyze the sign of \( f'(x) \):
- For \( x < \frac{1}{e} \), \( \log x < -1 \) so \( f'(x) < 0 \) (decreasing).
- For \( x > \frac{1}{e} \), \( \log x > -1 \) so \( f'(x) > 0 \) (increasing).

### Step 6: Evaluate \( f(x) \) at critical points
Now we evaluate \( f(x) \) at the critical point \( x = \frac{1}{e} \):
\[ f\left(\frac{1}{e}\right) = \frac{1}{e} \log\left(\frac{1}{e}\right) + 1 = \frac{1}{e} \cdot (-1) + 1 = 1 - \frac{1}{e} \]
Since \( 1 - \frac{1}{e} > 0 \) (as \( e > 2 \)), we have \( f\left(\frac{1}{e}\right) > 0 \).

### Step 7: Analyze behavior as \( x \to 0 \) and \( x \to \infty \)
- As \( x \to 0^+ \), \( f(x) \to 0 \) (since \( x \log x \to 0 \)).
- As \( x \to \infty \), \( f(x) \to \infty \) (since \( x \log x \to \infty \)).

### Step 8: Conclusion
From our analysis, we find that:
- \( f(x) > 0 \) for \( x \in \left(0, \frac{1}{e}\right) \cup \left(\frac{1}{e}, \infty\right) \).

Thus, the set \( \{ x : f(x) > 0 \} \) is:
\[ (0, \infty) \setminus \left\{ \frac{1}{e} \right\} \]

### Final Answer
The set \( \{ x : f(x) > 0 \} \) is equal to \( (0, \infty) \setminus \left\{ \frac{1}{e} \right\} \).

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