Find the points of local maxima/minima of function
`f(x) = x ln x`

Answer

To find the points of local maxima and minima of the function \( f(x) = x \ln x \), we will follow these steps:

### Step 1: Find the first derivative of \( f(x) \)
The first derivative \( f'(x) \) is calculated using the product rule. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative is given by:
\[ (uv)' = u'v + uv' \]
In our case, let \( u = x \) and \( v = \ln x \).

Calculating the derivatives:
- \( u' = 1 \)
- \( v' = \frac{1}{x} \)

Now applying the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:
\[ f'(x) = \ln x + 1 = 0 \]
Solving for \( x \):
\[ \ln x = -1 \]
Exponentiating both sides, we get:
\[ x = e^{-1} = \frac{1}{e} \]

### Step 3: Find the second derivative of \( f(x) \)
Now we need to find the second derivative \( f''(x) \) to determine the nature of the critical point. We differentiate \( f'(x) \):
\[ f'(x) = \ln x + 1 \]
The derivative of \( \ln x \) is \( \frac{1}{x} \), thus:
\[ f''(x) = \frac{1}{x} \]

### Step 4: Evaluate the second derivative at the critical point
Now we evaluate \( f''(x) \) at \( x = \frac{1}{e} \):
\[ f''\left(\frac{1}{e}\right) = \frac{1}{\frac{1}{e}} = e \]
Since \( e > 0 \), this indicates that the critical point \( x = \frac{1}{e} \) is a point of local minima.

### Conclusion
The function \( f(x) = x \ln x \) has a local minimum at the point \( x = \frac{1}{e} \).

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