Concaves 2

[Michael13]

Thanks for the help!!

Determine the intervals on which the following function is concave up or concave down.

g(x)=(3x+4)e^(-2x)

 

[HallsofIvy]

Do you know what "concave up and down" mean? What tests have you learned for concavity?

 

[Quaid]

Determine the intervals on which the following function is concave up or concave down.

g(x)=(3x+4)e^(-2x)

Hello, again.

Most of the exercises that you've posted today require similar steps to finish.

As a function's second derivative gives information about its concavity, the first steps in this exercise could be to determine the second derivative.

Have you done that, yet?

How do your class materials explain the Second Derivative Test for function concavity?

Please explain where you're at.

Cheers

 

[Michael13]

Sorry All.

Sorry all, I just wanted to throw them on here real quick.
This problem is the hardest for me as I have no idea how to get the first or second derivatives.
I think that f'=(-6)/(e^2x)-2(3/(e^2x)-(2(4+3x))/(e^2x))
I think that it would make f''=(-6)/(e^2x)-(2(4+3x))/(e^2x)
I have no idea please help!

 

[HallsofIvy]

I don't know why you are changing "\(\displaystyle e^{-2x}\)" to "\(\displaystyle \frac{1}{e^{2x}}\)". That only confuses things.

With \(\displaystyle g= (3x+ 4)e^{-2x}\) the product rule gives \(\displaystyle g'= (3x+ 4)' e^{-2x}+ (3x+ 4)(e^{-2x})'= 3e^{-2x}- 2(3x+ 4)e^{-2x}= (3- 6x- 8)e^{-2x}= (-6x- 5)e^{-2x}\).

The second derivative is \(\displaystyle g''= (-6x- 5)' e^{-2x}+(-6x- 5)(e^{-2x})'= -6e^{-2x}+ (-6x-5)(-2e^{-2x})= (-6+ 12x+ 10)e^{-2x}= (12x+ 4)e^{-2x}\).

 

[Michael13]

Thanks, so what I get from there, putting it all to zero, is e^-2x=0 which is no solution and x=-1/3.

when I plug 0 and -1 (numbers picked randomly on either side of the -1/3) I come up with g''(0)=4 and g''(-1)=-8e^2.

This gives me concave down at (-infinity, -.3333) and up at (-.3333, infinity)

how does this look?

 

[HallsofIvy]

Thanks, so what I get from there, putting it all to zero, is e^-2x=0 which is no solution and x=-1/3.

when I plug 0 and -1 (numbers picked randomly on either side of the -1/3) I come up with g''(0)=4 and g''(-1)=-8e^2.

This gives me concave down at (-infinity, -.3333) and up at (-.3333, infinity)

how does this look?

Do you have a reason for writing "-.33333" rather than "-1/3"? They are NOT the same, you know.

 

[Michael13]

just to make it pretty with decimals. I really thought they were the same... yikes!

with that in mind I think I am all set on this problem. Thanks a ton!!