Apps of Differentiation help

**FalconPUNCH!** · November 29th 2007, 05:40 PM

Need help finding (a) Critical Numbers and (b) Inflection points for these two

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

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

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

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

$y' = -e^x cosx$

$y'' = -e^x(-sinx)$ <--- 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

**FalconPUNCH!** · November 29th 2007, 06:55 PM

I also need help with when they concave down and up. I need help as soon as possible.

**kalagota** · November 29th 2007, 07:13 PM

Originally Posted by FalconPUNCH!

Need help finding (a) Critical Numbers and (b) Inflection points for these two

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

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

Originally Posted by FalconPUNCH!

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

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

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

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

Originally Posted by FalconPUNCH!

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

$y' = -e^x cosx$

$y'' = -e^x(-sinx)$ <--- 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 $y = e^{-x} \sin x$
so, $y' = -e^{-x} \sin x + e^{-x} \cos x$ and
$y'' = - (-e^{-x} \sin x + e^{-x} \cos x) + (-e^{-x} \cos x - e^{-x} \sin x) = -2e^{-x} \cos x$

**FalconPUNCH!** · November 29th 2007, 07:15 PM

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.

**kalagota** · November 29th 2007, 07:26 PM

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..

**FalconPUNCH!** · November 29th 2007, 07:30 PM

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

**FalconPUNCH!** · November 29th 2007, 07:54 PM

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 $\frac{pi}{2}$ * as a critical number. I'm not too sure if there's supposed to be another.

*It's probably wrong as well.

**kalagota** · November 29th 2007, 07:59 PM

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 $\frac{pi}{2}$ * 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 $x = \frac{\pi}{4}$

**FalconPUNCH!** · November 29th 2007, 08:02 PM

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.

**FalconPUNCH!** · November 29th 2007, 08:16 PM

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 $\frac{pi}{2}$ is a critical number though.

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

**kalagota** · November 29th 2007, 08:26 PM

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 $\frac{pi}{2}$ is a critical number though.

$sin(x) = 0$ ==> x = $\frac{pi}{2}$ 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..

**topsquark** · November 29th 2007, 08:32 PM

Originally Posted by FalconPUNCH!

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

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

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

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

$sin(x) = cos(x)$

$tan(x) = 1$

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

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

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

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

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

$cos(x) = 0$

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

-Dan

**FalconPUNCH!** · November 29th 2007, 08:37 PM

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 $\frac{pi}{4}$ because the interval is 0 < or equal to x < or equal to 2pi?

**FalconPUNCH!** · November 29th 2007, 08:49 PM

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.

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