Math Help - Apps of Differentiation help

1. Apps of Differentiation help

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

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

3. 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
$y'' = - (-e^{-x} \sin x + e^{-x} \cos x) + (-e^{-x} \cos x - e^{-x} \sin x) = -2e^{-x} \cos x$

4. 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.

5. 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..

6. 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

7. 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.

8. 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 $x = \frac{\pi}{4}$

9. 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.

10. 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

11. 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..

12. 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

13. 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?

14. 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.