9. [-/11.12 Points] DETAILS MY NOTES SCALCET9 4.1.059. Find the absolute maximum and absolute minimum values of \( f \) on the given \[ f(t)=t-\sqrt[3]{t},[-1,4] \] absolute minimum value absolute maximum value

To find the absolute maximum and minimum values of the function \( f(t) = t - \sqrt[3]{t} \) on the interval \([-1, 4]\), we first need to calculate the derivative \( f'(t) \) to identify critical points. The derivative is calculated as: \[ f'(t) = 1 - \frac{1}{3} t^{-\frac{2}{3}} \] Setting \( f'(t) = 0 \) allows us to solve for the critical points: \[ 1 - \frac{1}{3} t^{-\frac{2}{3}} = 0 \] \[ \frac{1}{3} t^{-\frac{2}{3}} = 1 \] \[ t^{-\frac{2}{3}} = 3 \] \[ t^{\frac{2}{3}} = \frac{1}{3} \] \[ t = \left(\frac{1}{3}\right)^{\frac{3}{2}} = \frac{1}{\sqrt{27}} \approx 0.1937 \] Now we evaluate \( f(t) \) at the critical point and the endpoints of the interval: 1. At \( t = -1: f(-1) = -1 - (-1)^{\frac{1}{3}} = -1 + 1 = 0 \). 2. At \( t \approx 0.1937: f(0.1937) \approx 0.1937 - (0.1937)^{\frac{1}{3}} \approx 0.1937 - 0.578 = -0.3843 \). 3. At \( t = 4: f(4) = 4 - \sqrt[3]{4} \approx 4 - 1.5874 = 2.4126 \). Comparing these values, we find: - Absolute minimum: \( f \approx -0.3843 \) at \( t \approx 0.1937 \). - Absolute maximum: \( f \approx 2.4126 \) at \( t = 4 \).