f=x^2+5x-14
f '=2x+5
f ''=2
since f'' > 0, always concave up (I mean, come on, it's a parabola!)
so, no inflections
this should have been a no-brainer
Let f (x) = x^2 + 5x-14. Find the open intervals on which f is concave up (down). Then determine the xcoordinates of all inflection points of f
1. f is concave up on the intervals
2. f is concave down on the intervals
3. The inflection points occur at x
1 answer