find second derivative

[ibysaiyan]

1. The problem statement, all variables and given/known data
A curve has equation y=e^2x-x^2+x-3 , find value of x for which d^2y/dx^2=0.


2. Relevant equations



3. The attempt at a solution
well. i started by finding out the 1st and 2nd derivative:
y=e^2x-x^2+x-3
dy/dx= 2e2^x-2x+1 and d2y/dx2=4e^2x-2 = 0

dy/dx =>2e^2x=2x-1
=>e^2x = 2x/2 -1/2
e2x= x-1/2 (1)

sub. value (1) into: d2y/dx2.
4e^2x-2= 0
e^2x = 1/2
e^2x = 1/2 (x-1/2)
no idea.. on what to do now =/.
Thanks in adv.

[ideasrule]

dy/dx =>2e^2x=2x-1
=>e^2x = 2x/2 -1/2
e2x= x-1/2 (1)


I don't know what you did here (why is 2e^2x=2x-1? dy/dx doesn't have to equal 0), but you already got 4e^2x-2 = 0 in the previous step. Just solve for x and you're done.

[Dick]

How did you go from e^(2x)=1/2, which looks ok, to e^(2x)=(1/2)*(x-1/2) which does not look ok? If it's after 4AM there, I suggest you take a nap.

[ibysaiyan]

Oh k, yea. i can barely hold my eyes lol, alright i guess i will sleep now, thanks for the replies and helping me out :), i will be back tomorrow.Good night for now.

[nesteel]

.....
try to use the inverse operation of the exponential

[Dick]

The policy of the forum is not to present solutions for problems, ok? Just give hints. Never do that again, ok? I'm not going to hit the Report button. But I will next time. ibysaiyan could have gotten this on his own. Don't you see the value in that?

[nesteel]

:[ sorry sorry, i won't do it again
...is it better now ? :D

[Dick]

:[ sorry sorry, i won't do it again
...is it better now ? :D

Much better, thanks!