# How do you find the absolute maximum and absolute minimum values of f on the given interval: f(t) =t sqrt(25-t^2) on [-1, 5]?

Jul 18, 2016

Reqd. extreme values are $- \frac{25}{2} \mathmr{and} \frac{25}{2}$.

#### Explanation:

We use substitution $t = 5 \sin x , t \in \left[- 1 , 5\right]$.

Observe that this substitution is permissible, because,

$t \in \left[- 1 , 5\right] \Rightarrow - 1 \le t \le 5 \Rightarrow - 1 \le 5 \sin x \le 5$

$\Rightarrow - \frac{1}{5} \le \sin x \le 1$,

which holds good, as range of $\sin$ fun. is $\left[- 1 , 1\right]$.

Now, $f \left(t\right) = t \sqrt{25 - {t}^{2}} = 5 \sin x \cdot \sqrt{25 - 25 {\sin}^{2} x}$

$= 5 \sin x \cdot 5 \cos x = 25 \sin x \cos x = \frac{25}{2} \left(2 \sin x \cos x\right) = \frac{25}{2} \sin 2 x$

Since, $- 1 \le \sin 2 x \le 1 \Rightarrow - \frac{25}{2} \le \frac{25}{2} \sin 2 x \le \frac{25}{2}$

$\Rightarrow - \frac{25}{2} \le f \left(t\right) \le \frac{25}{2}$

Therefore, reqd. extremities are $- \frac{25}{2} \mathmr{and} \frac{25}{2}$.

Jul 18, 2016

Find the monotony of the function from the derivative's sign and decide which local maximum/minimums are the biggest, smallest.

Absolute maximum is:
$f \left(3.536\right) = 12.5$

Absolute minimum is:
$f \left(- 1\right) = - 4.899$

#### Explanation:

$f \left(t\right) = t \sqrt{25 - {t}^{2}}$

The derivative of the function:

$f ' \left(t\right) = \sqrt{25 - {t}^{2}} + t \cdot \frac{1}{2 \sqrt{25 - {t}^{2}}} \left(25 - {t}^{2}\right) '$

$f ' \left(t\right) = \sqrt{25 - {t}^{2}} + t \cdot \frac{1}{2 \sqrt{25 - {t}^{2}}} \left(- 2 t\right)$

$f ' \left(t\right) = \sqrt{25 - {t}^{2}} - {t}^{2} / \sqrt{25 - {t}^{2}}$

$f ' \left(t\right) = {\sqrt{25 - {t}^{2}}}^{2} / \sqrt{25 - {t}^{2}} - {t}^{2} / \sqrt{25 - {t}^{2}}$

$f ' \left(t\right) = \frac{25 - {t}^{2} - {t}^{2}}{\sqrt{25 - {t}^{2}}}$

$f ' \left(t\right) = \frac{25 - 2 {t}^{2}}{\sqrt{25 - {t}^{2}}}$

$f ' \left(t\right) = 2 \frac{12.5 - {t}^{2}}{\sqrt{25 - {t}^{2}}}$

$f ' \left(t\right) = 2 \frac{{\sqrt{12.5}}^{2} - {t}^{2}}{\sqrt{25 - {t}^{2}}}$

$f ' \left(t\right) = 2 \frac{\left(\sqrt{12.5} - t\right) \left(\sqrt{12.5} + t\right)}{\sqrt{25 - {t}^{2}}}$

• The numerator has two solutions:
${t}_{1} = \sqrt{12.5} = 3.536$
${t}_{2} = - \sqrt{12.5} = - 3.536$
Therefore, the numerator is:
Negative for $t \in \left(- \infty , - 3.536\right) \cup \left(3.536 , + \infty\right)$
Positive for $t \in \left(- 3.536 , 3.536\right)$

• The denominator is always positive in $\mathbb{R}$, since it's a square root.
Finally, the range given is $\left[- 1 , 5\right]$

Therefore, the derivative of the function is:
- Negative for $t \in \left[- 1 , 3.536\right)$
- Positive for $t \in \left(3.536 , 5\right)$
This means the graph firstly goes up from $f \left(- 1\right)$ to $f \left(3.536\right)$ and then goes down to $f \left(5\right)$. This makes $f \left(3.536\right)$ the absolute maximum and the biggest value of $f \left(- 1\right)$ and $f \left(5\right)$ is the absolute minimum.

Absolute maximum is $f \left(3.536\right)$:
$f \left(3.536\right) = 3.536 \sqrt{25 - {3.536}^{2}} = 12.5$

For the absolute maximum:

$f \left(- 1\right) = - 1 \sqrt{25 - {\left(- 1\right)}^{2}} = - 4.899$

$f \left(5\right) = 5 \sqrt{25 - {5}^{2}} = 0$

Therefore, $f \left(- 1\right) = - 4.899$ is the absolute minimum.

You can see from the graph below that this is true. Just ignore the area left of $- 1$ since it's out of the domain:

graph{xsqrt(25-x^2) [-14.4, 21.63, -5.14, 12.87]}