Question

Find the absolute maximum and absolute minimum values of on the given interval.

43. \(f(x) = t\sqrt {4 - {t^2}} \), \(( - 1,\;2)\)

Short answer

The absolute maximum value of \(f(t)\) is \(f(\sqrt 2 ) = 2\) and the absolute minimum of \(f(t)\) is \(f( - 1) = - \sqrt 3 \).

Step-by-step solution

Given data

The function \(f(x) = t\sqrt {4 - {t^2}} \) in the interval, \(( - 1,\;2)\).

Concept of closed interval method

1. Find the values of\(f\)at the critical numbers of\(f\)in\((a,b)\)

2. Find the values of\(f\)at the endpoints of the interval.

3. The largest of the values from steps\(1\)and\(2\)is the absolute maximum value, the smallest of these values is the absolute minimum value.

Obtain the first derivative of the given function

Obtain the first derivative of the given function, \({f^\prime }(t) = \frac{d}{{dx}}\left( {t\sqrt {4 - {t^2}} } \right)\).

Apply the chain rule:

\(\begin{aligned}{c}{f^\prime }(x) &= \sqrt {4 - {t^2}} \frac{d}{{dx}}(t) + t\frac{d}{{dx}}\left( {\sqrt {4 - {t^2}} } \right)\\ &= \sqrt {4 - {t^2}} + t\left( { - \frac{t}{{\sqrt {4 - {t^2}} }}} \right)\\ &= \frac{{4 - {t^2} - {t^2}}}{{\sqrt {4 - {t^2}} }}\\ &= \frac{{4 - 2{t^2}}}{{\sqrt {4 - {t^2}} }}\end{aligned}\)

Find the critical number

Set \({f^\prime }(t) = 0\)and obtain the critical numbers.

\(\begin{aligned}{c}\frac{{4 - 2{t^2}}}{{\sqrt {4 - {t^2}} }} &= 0\\4 - 2{t^2} &= 0\\2{t^2} &= 4\\{t^2} &= 2\end{aligned}\)

Take square root of above equation and obtain, \(t = \pm \sqrt 2 \).

Thus, the critical numbers are \( - \sqrt 2 \) and \(\sqrt 2 \), which lies in the given interval \(( - 1,2)\).

Apply the extreme values of the given interval

Apply the extreme values of the given interval and the critical point values in \(f(t)\).

Substitute \(t = - 1\) in \(f(t)\).

\(\begin{aligned}{c}f( - 1) &= ( - 1)\sqrt {4 - {{( - 1)}^2}} \\ &= - \sqrt {4 - 1} \\ &= - \sqrt 3 \end{aligned}\)

Substitute \(t = - \sqrt 2 \) in \(f(t)\).

\(\begin{aligned}{c}f( - \sqrt 2 ) &= ( - \sqrt 2 )\sqrt {4 - {{( - \sqrt 2 )}^2}} \\ &= ( - \sqrt 2 )\sqrt {4 - 2} \\ &= ( - \sqrt 2 )\sqrt 2 \\ &= - 2\end{aligned}\)

Find the absolute maximum and absolute minimum of the given function \(f(x) = t\sqrt {4 - {t^2}} \)

Similarly as above, substitute \(t = \sqrt 2 \) in \(f(t)\).

\(\begin{aligned}{c}f(\sqrt 2 ) &= (\sqrt 2 )\sqrt {4 - {{(\sqrt 2 )}^2}} \\ &= (\sqrt 2 )\sqrt {4 - 2} \\ &= (\sqrt 2 )\sqrt 2 \\ &= 2\end{aligned}\)

Substitute \(x = 2\) in \(f(t)\).

\(\begin{aligned}{c}f(2) &= (2)\sqrt {4 - {{(2)}^2}} \\ &= 2\sqrt {4 - 4} \\ &= 2(0)\\ &= 0\end{aligned}\)

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of \(f(t)\) is \(2\)and the absolute minimum of \(f(t)\)is \( - \sqrt 3 \).

Therefore, the absolute maximum value of \(f(t)\) is \(f(\sqrt 2 ) = 2\) and the absolute minimum of \(f(t)\) is \(f( - 1) = - \sqrt 3 \).