# Apps of Differentiation help

• Nov 29th 2007, 04:40 PM
FalconPUNCH!
Apps of Differentiation help
Need help finding (a) Critical Numbers and (b) Inflection points for these two

1. $\displaystyle y = 2x - tanx$ ; $\displaystyle \frac{-pi}{2} < x < \frac{pi}{2}$

$\displaystyle y' = 2 - sec^2(x)$

$\displaystyle y'' = -sec(x)tan(x)$

2. $\displaystyle y = e^(-x) sinx$ ; 0 < or equal to x < or equal to 2pi

$\displaystyle y' = -e^x cosx$

$\displaystyle y'' = -e^x(-sinx)$ <--- Not sure if that's correct

I'm not very fond of trig functions and I need help with these two. I need help asap
• Nov 29th 2007, 05:55 PM
FalconPUNCH!
I also need help with when they concave down and up. I need help as soon as possible.
• Nov 29th 2007, 06:13 PM
kalagota
Quote:

Originally Posted by FalconPUNCH!
Need help finding (a) Critical Numbers and (b) Inflection points for these two

crtical numbers are the solutions of $\displaystyle f'(x) = 0$

the possible inflection points are the solution of $\displaystyle f''(x) = 0$
to check whether it is an inflection point, try to check the following:
let $\displaystyle x_o$ be a possible point of inflection.
if $\displaystyle f''(x< x_o)$ and $\displaystyle f''(x > x_o)$ are of different signs, then $\displaystyle (x_o, f(x_o))$ is an inflection point.
for concavity, if $\displaystyle x_o$ is a possible point of inflection, and if $\displaystyle f''(x > x_o) > 0$, then the fucntion is concave up on the interval $\displaystyle x > x_o$; if $\displaystyle f''(x > x_o) < 0$, then the fucntion is concave down on the interval $\displaystyle x > x_o$

Quote:

Originally Posted by FalconPUNCH!

1. $\displaystyle y = 2x - tanx$ ; $\displaystyle \frac{-pi}{2} < x < \frac{pi}{2}$

$\displaystyle y' = 2 - sec^2(x)$

$\displaystyle y'' = -sec(x)tan(x)$

so, $\displaystyle 2 - \sec ^2 x = 0 \implies 2 = \sec ^2 x \implies \sqrt 2 = \sec x \implies \cos x = \frac{\sqrt 2}{2} \implies x = \frac{\pi}{4}$..
and
$\displaystyle y'' = -2\sec ^2 x \, \tan x = 0 \implies \sec ^2 x = 0$ or $\displaystyle \tan x = 0, \implies x = 0$ (sec x is never zero)
Quote:

Originally Posted by FalconPUNCH!

2. $\displaystyle y = e^(-x) sinx$ ; 0 < or equal to x < or equal to 2pi

$\displaystyle y' = -e^x cosx$

$\displaystyle y'' = -e^x(-sinx)$ <--- Not sure if that's correct

I'm not very fond of trig functions and I need help with these two. I need help asap

i think, your function is supposed to be $\displaystyle y = e^{-x} \sin x$
so, $\displaystyle y' = -e^{-x} \sin x + e^{-x} \cos x$ and
$\displaystyle y'' = - (-e^{-x} \sin x + e^{-x} \cos x) + (-e^{-x} \cos x - e^{-x} \sin x) = -2e^{-x} \cos x$
• Nov 29th 2007, 06:15 PM
FalconPUNCH!
Well I know how to find them but with Trig Functions it's harder for me to do. Am I supposed to use radians? I'm not too sure.
• Nov 29th 2007, 06:26 PM
kalagota
Quote:

Originally Posted by FalconPUNCH!
Well I know how to find them but with Trig Functions it's harder for me to do. Am I supposed to use radians? I'm not too sure.

i've edited it.. did you see it?
hmm, yes, you have to do the angles in radians..
• Nov 29th 2007, 06:30 PM
FalconPUNCH!
Quote:

Originally Posted by kalagota
i've edited it.. did you see it?
hmm, yes, you have to do the angles in radians..

Oh I was unsure. Thanks for your help :)
• Nov 29th 2007, 06:54 PM
FalconPUNCH!
I'm having problems with the local max for the 2nd problem. I think there's another critical number I'm missing but I got $\displaystyle \frac{pi}{2}$ * as a critical number. I'm not too sure if there's supposed to be another.

*It's probably wrong as well.
• Nov 29th 2007, 06:59 PM
kalagota
Quote:

Originally Posted by FalconPUNCH!
I'm having problems with the local max for the 2nd problem. I think there's another critical number I'm missing but I got $\displaystyle \frac{pi}{2}$ * as a critical number. I'm not too sure if there's supposed to be another.

*It's probably wrong as well.

can you show us the solution..
i think, it was at max if $\displaystyle x = \frac{\pi}{4}$
• Nov 29th 2007, 07:02 PM
FalconPUNCH!
Quote:

Originally Posted by kalagota
can you show us the solution..

I would if I wrote it down. I'm doing it with a calculator right now but I'm going to try again. My trig teacher wasn't that great (was new at teaching) so I don't understand it fully because of the way he taught.
• Nov 29th 2007, 07:16 PM
FalconPUNCH!
Well I don't know what to do so I give up. I've looked in the book for help and asked someone else but no luck. Thanks for helping anyways.

I did get that $\displaystyle \frac{pi}{2}$ is a critical number though.

$\displaystyle sin(x) = 0$ ==> x = $\displaystyle \frac{pi}{2}$ not sure if it's correct though
• Nov 29th 2007, 07:26 PM
kalagota
Quote:

Originally Posted by FalconPUNCH!
Well I don't know what to do so I give up. I've looked in the book for help and asked someone else but no luck. Thanks for helping anyways.

I did get that $\displaystyle \frac{pi}{2}$ is a critical number though.

$\displaystyle sin(x) = 0$ ==> x = $\displaystyle \frac{pi}{2}$ not sure if it's correct though

to get the critical number, you have to equate the first derivative to 0, not the function itself..
• Nov 29th 2007, 07:32 PM
topsquark
Quote:

Originally Posted by FalconPUNCH!
2. $\displaystyle y = e^(-x) sinx$ ; 0 < or equal to x < or equal to 2pi

$\displaystyle y^{\prime} = -e^{-x}sin(x) + e^{-x}cos(x)$

For critical numbers we have
$\displaystyle -e^{-x}sin(x) + e^{-x}cos(x) = 0$

$\displaystyle -sin(x) + cos(x) = 0$

$\displaystyle sin(x) = cos(x)$

$\displaystyle tan(x) = 1$

This occurs when $\displaystyle x = \frac{\pi}{4}$ and $\displaystyle x = \frac{5 \pi}{4}$.

$\displaystyle y^{\prime \prime} = e^{-x}sin(x) - e^{-x}cos(x) - e^{-x}cos(x) + e^{-x} \cdot -sin(x)$

$\displaystyle y^{\prime \prime} = e^{-x}sin(x) - 2e^{-x}cos(x) - e^{-x}sin(x)$

$\displaystyle y^{\prime \prime} = -2e^{-x}cos(x)$

Inflection points are where $\displaystyle y^{\prime \prime} = 0$, so
$\displaystyle -2e^{-x}cos(x) = 0$

$\displaystyle cos(x) = 0$

This occurs when $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \frac{3 \pi}{2}$

-Dan
• Nov 29th 2007, 07:37 PM
FalconPUNCH!
I'm so stupid lol! I didn't set sinx = cosx and didn't know that I could do that. >_< I really need to go back and look over trig functions. Also I didn't even notice I was using the original function and not the derivative till you told me kalagota. Alright I'm gonna go back and finish the rest.

Tops, I don't add 360 to $\displaystyle \frac{pi}{4}$ because the interval is 0 < or equal to x < or equal to 2pi?
• Nov 29th 2007, 07:49 PM
FalconPUNCH!
Is it possible if one of you can give me a site with trig function identities and rules? I don't want to be a bother next time I come around them.