find the absolute minimum and absolute maximum values of f on the given interval.\n$f(t)=t\\sqrt{64…

Explanation:

Step1: Find the derivative of (f(t))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = t), (u^\prime=1), (v=\sqrt{64 - t^{2}}=(64 - t^{2})^{\frac{1}{2}}), and (v^\prime=\frac{-2t}{2\sqrt{64 - t^{2}}}=\frac{-t}{\sqrt{64 - t^{2}}}).

(f^\prime(t)=\sqrt{64 - t^{2}}+t\times\frac{-t}{\sqrt{64 - t^{2}}}=\frac{64 - t^{2}-t^{2}}{\sqrt{64 - t^{2}}}=\frac{64 - 2t^{2}}{\sqrt{64 - t^{2}}})

Step2: Find the critical points

Set (f^\prime(t) = 0), then (64-2t^{2}=0) (since the denominator (\sqrt{64 - t^{2}}\neq0) for (t\in[-1,8))).

(2t^{2}=64), (t^{2} = 32), (t=\pm4\sqrt{2}). But (t=- 4\sqrt{2}\approx - 5.66\notin[-1,8]), so (t = 4\sqrt{2}\approx5.66\in[-1,8])

Also, check the endpoints (t=-1) and (t = 8). When (t = 8), (f(8)=8\sqrt{64 - 64}=0)

Step3: Evaluate (f(t)) at critical points and endpoints

• (f(-1)=-1\times\sqrt{64-1}=-\sqrt{63}\approx - 7.94)
• (f(4\sqrt{2})=4\sqrt{2}\times\sqrt{64-(4\sqrt{2})^{2}}=4\sqrt{2}\times\sqrt{64 - 32}=4\sqrt{2}\times\sqrt{32}=4\sqrt{2}\times4\sqrt{2}=32)
• (f(8)=0)

Answer:

absolute minimum value: (-\sqrt{63}) (or approximately (-7.94))

absolute maximum value: (32)