Question

The minimum value of $x\log x$ is equal to

• A. $e$
• B. $1/e$
• C. $-1/e$
• D. $2/e$

Hint:

$\dfrac{d}{dx}(x\ln x) = \ln x + 1$. The minimum of $x\ln x$ occurs at $x = 1/e$ and equals $-1/e$.

Answer 1

The Correct Option is C

To find the minimum value of the function \( x \log x \), we need to use calculus, specifically the concept of derivatives, to identify the critical points and determine the minimum value. Let's go through the process step-by-step.

1. First, we need to define the function clearly. The function given is \(f(x) = x \log x\), which is defined for \(x > 0\) because the logarithm is only defined for positive real numbers.
2. To find the critical points, we need to find the derivative of the function and set it equal to zero. Using the product rule for derivatives, we have: \(f'(x) = \frac{d}{dx}(x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1\).
3. Set the derivative equal to zero to find critical points: \(\log x + 1 = 0 \Rightarrow \log x = -1\).
4. Solving for \(x\), we get: \(x = e^{-1} = \frac{1}{e}\).
5. To determine if this critical point is a minimum, we check the second derivative: \(f''(x) = \frac{d}{dx}(\log x + 1) = \frac{1}{x}\).
6. Since \(f''(x) = \frac{1}{x}\) is positive for \(x > 0\), the function \(x \log x\) is concave up at \(x = \frac{1}{e}\). Thus, this point is a minimum.
7. Substitute the critical point into the original function to find the minimum value: \(f\left(\frac{1}{e}\right) = \frac{1}{e} \log\left(\frac{1}{e}\right) = \frac{1}{e} (-1) = -\frac{1}{e}\).
8. Thus, the minimum value of \(x \log x\) is \(-\frac{1}{e}\), which corresponds to the option

$-1/e$

Hence, the correct answer is -1/e.