# Critical Numbers and Absolute Max/Min

• Nov 14th 2007, 02:39 PM
FalconPUNCH!
Critical Numbers and Absolute Max/Min
1) Find the Critical Numbers for
$h(p) = \frac{p - 1}{p^2+4}$

$h'(p) = \frac{p^2 + 4 - (p - 1)(2p)}{(p^2 + 4)^2}$

after all the work is done I get

$-p^2 + 2p + 4 = 0$

There aren't any critical numbers is what I got but I'm not sure.

2) Find the Critical Numbers for $g(x) = \sqrt{1 - x^2}$

$g'(x) = \frac{2x}{2 \sqrt{1-x^2}}$

$\frac{x}{\sqrt{1-x^2}} = 0$

I get $x = 0$ as a critical number but I'm not sure if it's right or the only answer

3) Find the Absolute max and min for
$f(t) = 2cos(t) + sin(2t)$ in the interval $[0 , \frac{pi}{2}]$

after simplifying the derivative for it I get:

$f'(t) = 2(cost(2t) + sin(t))$

I don't know what to do with it since I am bad with trig functions in these types of problems.
• Nov 14th 2007, 03:25 PM
Soroban
Hello, FalconPUNCH!

Here's #3 . . .

Quote:

3) Find the Absolute max and min for: . $f(t) \:= \:2\cos t + \sin2t$ in the interval $\left[0 ,\,\frac{\pi}{2}\right]$

The derivative is: . $f'(t) \:=\:-2\sin t + 2\cos2t$

We have: . $2\cos2t - 2\sin t \;=\;0$

. . Then: . $2\left(1-2\sin^2t\right) - 2\sin t \;=\;0$

. . which simplifies to: . $2\sin^2\!t + \sin t = 1\;=\;0$

. . which factors: . $(\sin t + 1)(2\sin t-1)\;=\;0$

Then we have: . $\sin t + 1 \:=\:0\quad\Rightarrow\quad\sin t \:=\:-1\quad\Rightarrow\quad t \:=\:\frac{3\pi}{2}$ . . .
not in the interval!

And we have: . $2\sin t - 1\:=\:0\quad\Rightarrow\quad \sin t\:=\:\frac{1}{2}\quad\Rightarrow\quad t \:=\:\frac{\pi}{6}$

Find the value of the function at the critical value and the endpoints:

$f(0) \;=\;2\cos(0) + \sin(0) \;=\;2\!\cdot\!1+0\;=\;2$

$f\left(\frac{\pi}{6}\right)\;=\;2\cos\left(\frac{\ pi}{6}\right) +\sin\left(\frac{\pi}{3}\right) \;=\;2\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\;\frac{3\sqrt{3}}{2} \;\approx\;2.598$ . ← absolute maximum

$f\left(\frac{\pi}{2}\right) \;=\;2\cos\left(\frac{\pi}{2}\right) + \sin(\pi) \;=\;2\!\cdot\!0 + 0 \;=\;0$ . ← absolute minimum

• Nov 14th 2007, 04:10 PM
FalconPUNCH!
Thank you. Can anyone check and make sure I didn't screw up on the first two?
• Nov 15th 2007, 07:50 AM
FalconPUNCH!
Can anyone help? I have school in an hour and I won't be back the whole day.
• Nov 15th 2007, 07:53 AM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
1) Find the Critical Numbers for
$h(p) = \frac{p - 1}{p^2+4}$

$h'(p) = \frac{p^2 + 4 - (p - 1)(2p)}{(p^2 + 4)^2}$

after all the work is done I get

$-p^2 + 2p + 4 = 0$

There aren't any critical numbers is what I got but I'm not sure.

this quadratic equation has roots, two in fact, and therefore, there are two critical numbers.
• Nov 15th 2007, 07:53 AM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
2) Find the Critical Numbers for $g(x) = \sqrt{1 - x^2}$

$g'(x) = \frac{2x}{2 \sqrt{1-x^2}}$

$\frac{x}{\sqrt{1-x^2}} = 0$

I get $x = 0$ as a critical number but I'm not sure if it's right or the only answer

that is correct (this is the upper have of the circle with radius 1 centered at the origin. we have a horizontal slope at the top)
• Nov 15th 2007, 08:24 AM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
this quadratic equation has roots, two in fact, and therefore, there are two critical numbers.

So do I just use the quadratic formula since I can't factor it?