Find the absolute minimum and absolute maximum values of \( f \) on the given interval. \[ f(t)=t \sqrt{25-t^{2}}, \quad[-1,5] \] absolute minimum \( \square \)

To find the absolute minimum and maximum values of \( f(t) = t \sqrt{25 - t^2} \) on the interval \([-1, 5]\), we first find the derivative of \( f \). Using the product and chain rules, we have: \[ f'(t) = \sqrt{25 - t^2} + t \cdot \frac{-t}{\sqrt{25 - t^2}} = \frac{25 - 2t^2}{\sqrt{25 - t^2}}. \] Setting \( f'(t) = 0 \) gives us: \[ 25 - 2t^2 = 0 \implies t^2 = \frac{25}{2} \implies t = \pm \frac{5\sqrt{2}}{2}. \] Calculate \( t \) within the interval \([-1, 5]\): - \( \frac{5\sqrt{2}}{2} \approx 3.54 \) (valid) - \( -\frac{5\sqrt{2}}{2} \) is outside the interval. Now evaluate \( f(t) \) at the critical point and endpoints: 1. At \( t = -1 \): \[ f(-1) = -1 \sqrt{25 - 1} = -1 \cdot \sqrt{24} \approx -4.89. \] 2. At \( t = 5 \): \[ f(5) = 5 \sqrt{25 - 25} = 0. \] 3. At \( t = \frac{5\sqrt{2}}{2} \): \[ f\left(\frac{5\sqrt{2}}{2}\right) = \frac{5\sqrt{2}}{2} \sqrt{25 - \frac{25}{2}} = \frac{5\sqrt{2}}{2} \cdot \sqrt{\frac{25}{2}} = \frac{5\sqrt{2}}{2} \cdot \frac{5}{\sqrt{2}} = \frac{25}{2} = 12.5. \] Now we can summarize: - \( f(-1) \approx -4.89 \) - \( f(5) = 0 \) - \( f\left(\frac{5\sqrt{2}}{2}\right) = 12.5 \) Thus, the absolute minimum value is approximately \( -4.89 \) and the absolute maximum value is \( 12.5 \). Absolute minimum: \(-4.89\) (approximately) Absolute maximum: \(12.5\)