# Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = t -...

## Question:

Find the absolute maximum and absolute minimum values of {eq}f {/eq} on the given interval.

{eq}\displaystyle f(t) = t - \sqrt[3]{t}, \quad ~[-1,6] {/eq}

## Absolute Maxima and Absolute Minima:

To evaluate the absolute maxima and absolute minima of the function we 'll first calculate the critical points of the function. Next, we substitute the critical points and the endpoints of the interval to the original function to find the absolute maxima and absolute minima values.

## Answer and Explanation:

We have to find the absolute maximum and absolute minimum values of {eq}f {/eq} on the given interval.

{eq}\displaystyle f(t) = t - \sqrt[3]{t}, \quad ~[-1,6] {/eq}

First, we calcultae the critical points of the function.

Differentiate the given function with respect to t.

{eq}\begin{align} f'(t) & =\frac{\mathrm{d} }{\mathrm{d} t}( t - \sqrt[3]{t})\\[0.2cm] &=1-\frac{1}{3}t^{\frac{1}{3}-1}\\[0.2cm] &=1-\frac{1}{3}t^{\frac{-2}{3}}\\[0.2cm] &=1-\frac{1}{3t^{\frac{2}{3}}} \end{align} {/eq}

Set the derivative equal to zero.

{eq}\begin{align} f'(t) &=0\\ 1-\frac{1}{3t^{\frac{2}{3}}} &=0\\[0.2cm] \frac{3t^{\frac{2}{3}-1}}{t^{\frac{2}{3}}} &=0\\[0.2cm] 3t^{\frac{2}{3}}-1 &=0\\[0.2cm] 3t^{\frac{2}{3}} &=1\\[0.2cm] t^{\frac{2}{3}} &=\frac{1}{3}\\[0.2cm] t &=\frac{1}{3^{\frac{3}{2}}} \end{align} {/eq}

Now, we substitute the critical points and the endpoints of the interval to the original function.

{eq}f\left ( \dfrac{1}{3^{\frac{3}{2}}} \right ) = \dfrac{1}{3^{\frac{3}{2}}} - \sqrt[3]{\dfrac{1}{3^{\frac{3}{2}}}}=\dfrac{1}{3^{\frac{3}{2}}}-\dfrac{1}{3^{\frac{1}{2}}} \approx -0.3849\ \left ( Absolute\ Minima \right )\\[0.2cm] f\left ( -1 \right )=-1-\sqrt[3]{-1}=0\\[0.2cm] f(6)=6-\sqrt[3]{6} \approx 4.1829 \left ( Absolute\ Maxima \right ) {/eq}

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from General Studies Math: Help & Review

Chapter 5 / Lesson 2