the graph of is a parabola that opens upward.
says the graph of is concave up everywhere.
Hi guys, so I'm having trouble with the following question:
For f'(x), I got 2x and for f"(x) I got 2...
I need the values for which f"(x)=0, but how can I find it then f"(x)=2? I couldn't find any similar examples on Google guys, please help.
Thanks i advance.
Cheers
Theres an interval where f(x) is concave down, but I cant seem to find it also...
I tought it would never be concave down...
You can say that is concave up everywhere just because 2 > 0?
EDIT:
For A), the interval is (-inf,inf), but B) insists that there's still an interview where the function is concave down... I dont understand how is that possible since it is concave up everywhere...
There is no interval where is concave down. Some advice ...
(1) LOOK at the graph of . Frankly, I'm surprised that a calculus student cannot sketch this simple graph from memory.
(2) Review your notes/text over what the sign of says about the graph of .
I got the interval where it is concave up. I could sketch this graph and find the interval thanks to your help. But the question still asks for an interval where the function is concave down. its not cocave down anywhere, but it wont accept my answer. Ive tried to put "none" or "0" and it doesnt work... Maybe a problem with the website?
Thanks mate!
Since the second derivative is positive for all real x, then the function is concave up for all real x. I can't help with what to enter though, since I went to school in the dark ages where a professor actually manually graded homework that was turned in by hand.