# Find second derivative

1. Dec 8, 2009

### 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 =/.

2. Dec 8, 2009

### ideasrule

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.

3. Dec 8, 2009

### 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.

4. Dec 8, 2009

### 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.

5. Dec 8, 2009

### nesteel

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

Last edited: Dec 8, 2009
6. Dec 8, 2009

### 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?

Last edited: Dec 8, 2009
7. Dec 8, 2009

### nesteel

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

8. Dec 8, 2009

### Dick

Much better, thanks!