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

Explanation:

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

The function is (f(t)=t - t^{\frac{1}{3}}). Using the power rule ((x^n)^\prime=nx^{n - 1}), the derivative (f^\prime(t)=1-\frac{1}{3}t^{-\frac{2}{3}}=1-\frac{1}{3t^{\frac{2}{3}}}). 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^{2}&=\frac{1}{27}\ t&=\pm\frac{1}{3\sqrt{3}}=\pm\frac{\sqrt{3}}{9}\approx\pm0.192 \end{align*} ] Also, (f^\prime(t)) is undefined when (t = 0) (since the denominator (3t^{\frac{2}{3}}=0) at (t = 0)).

Step2: Evaluate (f(t)) at critical points and endpoints

• For (t=-1): (f(-1)=-1-(-1)^{\frac{1}{3}}=-1 + 1=0).
• For (t = 0): (f(0)=0-0^{\frac{1}{3}}=0).
• For (t=\frac{\sqrt{3}}{9}): (f(\frac{\sqrt{3}}{9})=\frac{\sqrt{3}}{9}-\left(\frac{\sqrt{3}}{9}\right)^{\frac{1}{3}}). Let (x=\frac{\sqrt{3}}{9}\approx0.192), (f(x)=x - x^{\frac{1}{3}}\approx0.192-0.577=-0.385).
• For (t = 6): (f(6)=6-6^{\frac{1}{3}}\approx6 - 1.817 = 4.183).

Answer:

absolute minimum value: (-\frac{2}{3\sqrt{3}}\approx - 0.385) absolute maximum value: (6-\sqrt[3]{6}\approx4.183)