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.

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

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

solving it algebraically i mean

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. >_<

do you know how to find the derivative?

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

ok now graph your 2nd derivative in your calculator

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

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

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

okay i got it now thanks!

yes right

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

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