Separate the interval of convaity of y =x `log_(e)x-(x^(2))/(2)+(1)/(2)`

Explanation

To find the intervals of concavity for the function \( y = x \log_e x - \frac{x^2}{2} + \frac{1}{2} \), we will follow these steps:

### Step 1: Find the first derivative \( y' \)

The first derivative of the function is obtained using the product rule and the power rule.

\[ y' = \frac{d}{dx}(x \log_e x) - \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}\left(\frac{1}{2}\right) \]

Calculating each term:
- The derivative of \( x \log_e x \) is \( 1 + \log_e x \) (using the product rule).
- The derivative of \( -\frac{x^2}{2} \) is \( -x \).
- The derivative of \( \frac{1}{2} \) is \( 0 \).

Thus, we have:

\[ y' = 1 + \log_e x - x \]

### Step 2: Find the second derivative \( y'' \)

Now, we differentiate \( y' \) to find \( y'' \):

\[ y'' = \frac{d}{dx}(1 + \log_e x - x) \]

Calculating each term:
- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( \log_e x \) is \( \frac{1}{x} \).
- The derivative of \( -x \) is \( -1 \).

Thus, we have:

\[ y'' = \frac{1}{x} - 1 \]

### Step 3: Set the second derivative to zero to find inflection points

To find the points of concavity change, we set \( y'' = 0 \):

\[ \frac{1}{x} - 1 = 0 \]

Solving for \( x \):

\[ \frac{1}{x} = 1 \implies x = 1 \]

### Step 4: Determine the sign of \( y'' \) in the intervals

We will test the sign of \( y'' \) in the intervals \( (0, 1) \) and \( (1, \infty) \):

1. For \( x \in (0, 1) \):
- Choose \( x = 0.5 \):
\[ y''(0.5) = \frac{1}{0.5} - 1 = 2 - 1 = 1 \quad (\text{positive}) \]
This implies that the function is concave upward in this interval.

2. For \( x \in (1, \infty) \):
- Choose \( x = 2 \):
\[ y''(2) = \frac{1}{2} - 1 = 0.5 - 1 = -0.5 \quad (\text{negative}) \]
This implies that the function is concave downward in this interval.

### Conclusion

The intervals of concavity for the function \( y = x \log_e x - \frac{x^2}{2} + \frac{1}{2} \) are:
- Concave upward on \( (0, 1) \)
- Concave downward on \( (1, \infty) \)

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