Determine the intervals where the function \( f(x) = \frac{\log(2x)}{x} \) is increasing and decreasing.
Find the derivative of \( f(x) \).
To find the derivative of \( f(x) = \frac{\log(2x)}{x} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = \log(2x) \) and \( v = x \). The derivative of \( u \), \( u' = \frac{1}{2x} \cdot 2 = \frac{1}{x} \), and the derivative of \( v \), \( v' = 1 \). Therefore, the derivative \( f'(x) = \frac{\frac{1}{x} \cdot x - \log(2x) \cdot 1}{x^2} = \frac{1 - \log(2x)}{x^2} \).
Determine where \( f'(x) > 0 \) (increasing) and \( f'(x) < 0 \) (decreasing).
Since \( x^2 \) is always positive for \( x \) in the domain, the sign of \( f'(x) \) depends on the sign of \( 1 - \log(2x) \). For \( f'(x) > 0 \), we need \( 1 - \log(2x) > 0 \), which implies \( \log(2x) < 1 \). This means \( 2x < e \), so \( x < \frac{e}{2} \). Also, since the logarithm is defined only for positive arguments, \( 2x > 0 \) implies \( x > 0 \). Therefore, \( f(x) \) is increasing on the interval \( (0, \frac{e}{2}) \). For \( f'(x) < 0 \), we need \( 1 - \log(2x) < 0 \), which implies \( \log(2x) > 1 \). This means \( 2x > e \), so \( x > \frac{e}{2} \). Therefore, \( f(x) \) is decreasing on the interval \( (\frac{e}{2}, \infty) \).
\(\boxed{\text{Increasing on } (0, \frac{e}{2}) \text{ and decreasing on } (\frac{e}{2}, \infty)}\)
\(\boxed{\text{Increasing on } (0, \frac{e}{2}) \text{ and decreasing on } (\frac{e}{2}, \infty)}\)
Answered
