# Thread: Critical Numbers and Absolute Max/Min

1. ## Critical Numbers and Absolute Max/Min

1) Find the Critical Numbers for
$\displaystyle h(p) = \frac{p - 1}{p^2+4}$

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

after all the work is done I get

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

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

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

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

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

I get $\displaystyle x = 0$ 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
$\displaystyle f(t) = 2cos(t) + sin(2t)$ in the interval $\displaystyle [0 , \frac{pi}{2}]$

after simplifying the derivative for it I get:

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

I don't know what to do with it since I am bad with trig functions in these types of problems.

2. Hello, FalconPUNCH!

Here's #3 . . .

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

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

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

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

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

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

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

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

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

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

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

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

3. Thank you. Can anyone check and make sure I didn't screw up on the first two?

4. Can anyone help? I have school in an hour and I won't be back the whole day.

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

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

after all the work is done I get

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

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.

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

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

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

I get $\displaystyle x = 0$ 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)

7. 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?