The graph of function y=x^3+6x^2+7x-2cos x, changes concavity at x=?

Could someone please help me with this math problem, I'm having a hard time understanding how to get the correct answer.

Question

The graph of function y=x^3+6x^2+7x-2cos x, changes concavity at x=?

Could someone please help me with this math problem, I'm having a hard time understanding how to get the correct answer.

myininaya · Answer

$$y'=3x^2+12x+7+2\sin(x) =>y''=6x+12+0+2\cos(x)$$
$$y''=0=>6x+12+2\cos(x)=0=>6x+2\cos(x)=-12$$

myininaya · Answer

i suggest using a graphing tool to find where the concavity changes. 
solving 6x+2cos(x)=-12 is not possible i believe

myininaya · Answer

solving it algebraically i mean

anonymous · Answer

Could you explain how to solve it with a graphing calculator.  I have one that I can use, it's just I'm not sure what steps to use on the calculator to go about solving it.  >_<

myininaya · Answer

do you know how to find the derivative?

anonymous · Answer

yes i found the 1st and 2nd derivative i'm just not sure what to do after that.

myininaya · Answer

ok now graph your 2nd derivative in your calculator

myininaya · Answer

where your graph intersects the x-axis is where you have inflection points

anonymous · Answer

and the inflection points are where the concavity changes, right?

anonymous · Answer

and the inflection points are where the concavity changes, right?

anonymous · Answer

okay i got it now thanks!

myininaya · Answer

yes right

anonymous · Answer

Thank you again for all the help, I very much appreciated it!!