Chapter 4.2, Problem 47E

To determine

To find: The absolute maximum and minimum value of given function.

Answer to Problem 47E

The absolute maximum and minimum value of given function is $2$ and $-\sqrt{3}$ respectively.

Explanation of Solution

Given information: The function is $f\left(t\right)=t\sqrt{4-t^2}$ and the interval is $\left[-1,2\right]$ .

Calculation:

Consider the function.

  $f\left(t\right)=t\sqrt{4-t^2}$ .................. (1)

Differentiate above equation.

  $\begin{array}{c}{f}^{\prime }\left(t\right)=\sqrt{4-t^2}+t\left(-\frac{2t}{2\sqrt{4-t^2}}\right)\\ =\sqrt{4-t^2}-\frac{t}{\sqrt{4-t^2}}\end{array}$

Substitute $f^{\prime }\left(t\right)=0$ in above equation.

  $\begin{array}{l}\sqrt{4-t^2}-\frac{t}{\sqrt{4-t^2}}=0\\ 4-t^2-t^2=0\\ 4-2t^2=0\\ t^2=2\\ t=\pm \sqrt{2}\end{array}$

Substitute the values in Equation (1).

  $\begin{array}{c}f\left(\sqrt{2}\right)=\sqrt{2}\sqrt{4-{\left(\sqrt{2}\right)}^2}\\ =2\\ f\left(-1\right)=\left(-1\right)\sqrt{4-{\left(-1\right)}^2}\\ =-\sqrt{3}\\ f\left(2\right)=\left(2\right)\sqrt{4-{\left(2\right)}^2}\\ =0\end{array}$

Thus, absolute maximum and minimum value of given function is $2$ and $\left(-1\right)\sqrt{4-{\left(-1\right)}^2}$ $-\sqrt{3}$ respectively