Question

f(x) is continuous for all x and has a local maximum at (1,8) which must be true
f'(2)=0
f' exists at x=2
the graph is concave down at x=1
f'(x) <0 if x<1 , f'(x) >0 if x > 1
f'(x) >0 if x <1 , f'(x) < 0 if x >1

Answer 1

It is concave down at x=1 and f'(x) > 0 if x < 1, f'(x) < 0 if x > 1

Answer 2

The graph does NOT have to be concave down at x = 1, but the other part of the previous answer is correct.

Answer 3

Oh god thats true, I was thinking about global maximums lol. Sorry

Answer 4

and functions of degree more than one :P

Answer 5

Global maximums don't have to be concave down either. There could be a cusp. I tried to use the drawing tool to show it, but I couldn't get it to work.

Answer 6

If there is a cusp, is it not undefined there?

Answer 7

f' is undefined there, but f is defined.

Answer 8

Yes. Thanks for correction, maria =)

Answer 9

That's why they pay me the big bucks.

Answer 10

Hopefully that's why they'll pay be big bucks in a few years too ^^

Answer 11

me*

Answer 12

Don't count on it, but good luck!

Answer 13

But, the function was supposed to be continous, maria. Thus no cusp. But anyway, since its a local max it need not be concave down there. It could even have been a linear expression as well.

Answer 14

NO. A continuous function can have a cusp. The function is continuous at a cusp, but the derivative isn't continuous there. The problem did not specify that the derivative had to be continuous.

Answer 15

Linear functions won't have local max or mins. You are thinking of absolute max or mins, on a specified interval.

Answer 16

Whats with me this morning...? Humiliating. I know this stuff, or usually I did anyway lol.

Answer 17

There are lots of ways to state things and it can get confusing. Don't worry about it. In real life you know what type of thing you are looking for, so the terminology won't mess you up.