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November 28, 2014

November 28, 2014

Posted by **Kris** on Sunday, May 6, 2012 at 12:17pm.

My answer for h'(x) is 2x - ksin(kx) but I'm not sure if that's right.

Then the answer I got for h''(x) was h''(x) = 2 - k^2cos(kx) but I'm pretty sure that's wrong based on the graph I was given.

Is my answer for the first derivative right? Thanks!

- Calculus AB -
**Kris**, Sunday, May 6, 2012 at 12:31pmOh nevermind, something was just set weird on my calculator that was throwing the graph off. I reset it and it looked like it was supposed to.

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