Question

The function f(x) = x4 – 6x2 + 8x – 3 has critical values at x = –2 and x = 1.

Determine
whether each of these is a maximum, a minimum or a point of inflection.

Answer 1

2nd derivative test

Answer 2

ok so plug in for -2 and for the second deriv and the same for 1 ? i got a negative number for -2 and 0 for 1

Answer 3

f''(c)<0, maximum
f''(c)>0, minimum

Answer 4

oh so it doesnt matter if they are negative and 0?

Answer 5

if its negative, its less than 0, so its a max

Answer 6

if its zero, doesn't really say anything other than its a possible point of inflection

Answer 7

concave up to concave down or vice versa

Answer 8

oh so its an inflection and no min

Answer 9

f'(x)=4x^3-12x+8
f''(x)=12x^2-12
f''(-2)=36
f''(1)=0
where'd the negative come from?

Answer 10

oh ok so its a min and inflection w/ no max value

Answer 11

woops a negative

Answer 12

i think -2 is a min and 1 is an inflextion point, but i could be wrong

Answer 13

correct

Answer 14

man i cant make those mistakes like that i got a exam tomorrow on this

Answer 15

to test for inflection, use values close to x=1, like x=0.9 and x=1.1. evaluate the 2nd derivative at those points, if it goes from neg to pos or pos to neg it's an inflection point

Answer 16

that is assuming of course that f''(c)=0 at that critical point

Answer 17

a negative could of killed the right answer

Answer 18

well at least you figured out what you did wrong before it's too late

Answer 19

what is that called the table of signs?

Answer 20

what is what called?

Answer 21

to test for inflection, use values close to x=1, like x=0.9 and x=1.1. evaluate the 2nd derivative at those points, if it goes from neg to pos or pos to neg it's an inflection point

Answer 22

yeah using the 2nd derivative will tell you if its concave up or down at those areas. if it goes concave down to up or up to down, its a point of inflection. only doing that because f''(1)=0, which doesn't tell us much except that theres a possible point of inflection