find the absolute maximum and absolute minimum values of f on the given interval. f(t)=t - \\sqrt3{t}, -1, 3…

Explanation:

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

Rewrite (f(t)=t - t^{\frac{1}{3}}). Using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (f^\prime(t)=1-\frac{1}{3}t^{-\frac{2}{3}}=1-\frac{1}{3t^{\frac{2}{3}}}).

Step2: Find the critical points

Set (f^\prime(t) = 0), then (1-\frac{1}{3t^{\frac{2}{3}}}=0). [ \begin{align*} \frac{1}{3t^{\frac{2}{3}}}&=1\ 3t^{\frac{2}{3}}&=1\ t^{\frac{2}{3}}&=\frac{1}{3}\ t&=\pm\frac{1}{3\sqrt{3}} \end{align*} ] Also, (f^\prime(t)) is undefined when (t = 0). So the critical points are (t=0,t=\frac{1}{3\sqrt{3}},t =-\frac{1}{3\sqrt{3}}). All of these critical points are in the interval ([-1,3]).

Step3: Evaluate the function at critical points and endpoints

• (f(-1)=-1-(-1)^{\frac{1}{3}}=-1 + 1=0).
• (f(0)=0-0^{\frac{1}{3}}=0).
• (f(\frac{1}{3\sqrt{3}})=\frac{1}{3\sqrt{3}}-(\frac{1}{3\sqrt{3}})^{\frac{1}{3}}). Let (x=\frac{1}{3\sqrt{3}}=\frac{1}{3\times3^{\frac{1}{2}}}=3^{-\frac{3}{2}}), then (f(\frac{1}{3\sqrt{3}})=3^{-\frac{3}{2}}-3^{-\frac{1}{2}}=3^{-\frac{1}{2}}(3^{-1}-1)=-\frac{2}{3\sqrt{3}}\approx - 0.38).
• (f(3)=3-\sqrt[3]{3}\approx3 - 1.44 = 1.56).

Answer:

absolute minimum value: (-\frac{2}{3\sqrt{3}}) absolute maximum value: (3-\sqrt[3]{3})