Question

need help on the attachment

Answer 1

guy answered it last time this is a repeat your local maximums will be located when f'(x) = 0 so solving the equation (2x^2+2x-12)(1+g(x)^2) = (2x+6)(x-2)(1+g(x)^2)

solving each equation will for 0 since if any of these equations go to 0, the the derivative will be zero you get
x=-3
x=2

the last one is one you must think about

(1+g(x)^2)=0
we know that g(x)^2 must be -1 to make this zero however you will always get a positive value due to squaring whatever that number must be so your critical numbers are -3 , 2 (these are where your local maximum and minimums will occur

Answer 2

does this help ta123?

Answer 3

so there just local max at x=-2 and local min at x=3

Answer 4

check your calculator put in 2x^2+2x-12 and look for zeroes to be 100%

Answer 5

yeah the x-intercept are x=-3, 2, so local max is x=-3 and local min is x=2

Answer 6

not quite your critical numbers are x=-3 and x=2... or the points at which your local max and min occur. the questions asks for points correct?

Answer 7

its asking for points where f will have local maximum

Answer 8

did you use the first derivative test

Answer 9

you mean plugging in test point in 4X+2

Answer 10

does it give you the actual function of the derivative? or just the derivative?

Answer 11

gives you the derivative

Answer 12

i mean like picking points on each side of the critical to see whether you get - or positive

Answer 13

no I just pick random points

Answer 14

do you know where it maximum at?

Answer 15

pick for example -2 , -4 and solve it,

Answer 16

what I think you meant maximum -3 and minimum is 2

Answer 17

ughh let me actually write this down i'm doing this in my head and it's confusign

Answer 18

yes -3 = maximum 2 = minimum

need some paper =/

Answer 19

Thanks!

Answer 20

yeah sorry about all that haha

Answer 21

that ok