The minimum value of f(x) = 2x - log x , x gt 0 , is

The correct answer is Option (3) → \(1 + \log 2\)

Setting the derivative equal to zero:

    \( \displaystyle f'(x) = 2 - \frac{1}{x} = 0 \), so \( \displaystyle x = \frac{1}{2} \)

Since \( \displaystyle f''(x) = \frac{1}{x^{2}} \gt 0 \), this is a minimum:

    \( \displaystyle f\left(\frac{1}{2}\right) = 1 - \log\frac{1}{2} \)

Using \( \log\frac{1}{2} = -\log 2 \):

    \( \displaystyle f\left(\frac{1}{2}\right) = 1 + \log 2 \)

This is the only critical point, and f decreases before it and increases after it, so \(1 + \log 2\) is the minimum value.