Given \( f(x)=x^{2}+\frac{16}{x},[1,4] \) Find(a) critical points (b) absolute maxi and absolute min

To find critical points and absolute maxima and minima for the function \( f(x) = x^{2} + \frac{16}{x} \) on the interval \([1, 4]\), we start by finding the derivative: 1. Find \( f'(x) \): \[ f'(x) = 2x - \frac{16}{x^2} \] 2. Set the derivative equal to zero to find the critical points: \[ 2x - \frac{16}{x^2} = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2 \] 3. Check that \( x = 2 \) lies in the interval \([1, 4]\). Next, evaluate \( f(x) \) at the critical point and at the endpoints of the interval: - Endpoints: \[ f(1) = 1^2 + \frac{16}{1} = 1 + 16 = 17 \] \[ f(4) = 4^2 + \frac{16}{4} = 16 + 4 = 20 \] - Critical point: \[ f(2) = 2^2 + \frac{16}{2} = 4 + 8 = 12 \] Finally, summarize the values: - \( f(1) = 17 \) - \( f(2) = 12 \) - \( f(4) = 20 \) Thus, the critical point is \( x = 2 \), the absolute minimum value is \( 12 \) at \( x = 2 \), and the absolute maximum value is \( 20 \) at \( x = 4 \).