f'' > 0 => concave up.
f''(x) = 34+12x^2
since f'' is always positive, f is always concave up.
you can see that because x^2 and x^4 are always positive, and always increasing when x is nonzero.
also, f' = 2x(2x^2+17) which is zero only at x=0. So, only one turning point at (0,0)
Determine the intervals on which the function is concave up and concave down
f(x) = 17x^2 = x^4
by differentiating I get
f'(x) = 34x+4x^3
and then I get lost. Can someone finish this up?
1 answer