Determine the maximum and minimum values of the function f(x)=-2log_(2)x on the interval [2,4]

To determine the maximum and minimum values of the function \( f(x) = -2\log_{2}x \) on the interval \([2, 4]\), we need to evaluate the function at critical points within the interval and at the endpoints.<br /><br />First, let's find the derivative of the function to locate any critical points:<br /><br />\[ f(x) = -2\log_{2}x = -2 \cdot \frac{\ln x}{\ln 2} \]<br /><br />The derivative is:<br /><br />\[ f'(x) = -2 \cdot \frac{1}{x \ln 2} = -\frac{2}{x \ln 2} \]<br /><br />Setting the derivative equal to zero to find critical points:<br /><br />\[ -\frac{2}{x \ln 2} = 0 \]<br /><br />This equation has no solutions because \(-\frac{2}{x \ln 2}\) is never zero for \(x > 0\). Therefore, there are no critical points in the interval \((2, 4)\).<br /><br />Next, evaluate the function at the endpoints of the interval:<br /><br />1. At \(x = 2\):<br /> \[<br /> f(2) = -2\log_{2}2 = -2 \times 1 = -2<br /> \]<br /><br />2. At \(x = 4\):<br /> \[<br /> f(4) = -2\log_{2}4 = -2 \times 2 = -4<br /> \]<br /><br />Comparing these values, we see that:<br /><br />- The maximum value of \(f(x)\) on \([2, 4]\) is \(-2\) at \(x = 2\).<br />- The minimum value of \(f(x)\) on \([2, 4]\) is \(-4\) at \(x = 4\).