Find the values of C such that the graph of F(x)=x^4+2x^3+Cx^2+2x+2 is concave upward everywhere

Question

jonnyvonny · Answer

Are you in calculus?

anonymous · Answer

Yes. I know the answer is c is greater than or equal to 1.5. I found the first and second derivative. When I factored the 2nd derivative using the QF, I got to the point -6 +- sqrt 36-24c /12. So then I set 36-24c as greater than or = to 0, since it's under the radical sign. But I'm getting c is LESS THAN or equal to zero!

jonnyvonny · Answer

Mmk, lemme have a crack at it.

anonymous · Answer

I'm also getting f"(x) = -1/2, and for f(x) to be concave upward, f"(x) must be positive.

jonnyvonny · Answer

You can take the third derivative:$$f'''(x)=24x+12\rightarrow f'''(x)=12(2x+1)$$
From here, we know if the third derivative is positive, it is concave up; the values of this domain which make it positive determines whether the graph is concave up or down.

jonnyvonny · Answer

That's how I would do it, however, looking at the answer...>.>

anonymous · Answer

We haven't done third derivatives (although I know how) but how does that give me the answer c is greater than or equal to 1.5?

jonnyvonny · Answer

Well, when you get x alone, you'll get x>1/2 is when its concave up.

jonnyvonny · Answer

However, you do the QF for the second derivative, how did you simplify what's under the radical?|dw:1386195683886:dw|