Question

need help on the attachment

Answer 1

is it the same one

Answer 2

well if you could go back to the old post because I wasn't getting negative for 3

Answer 3

alright so i found out why it was becoming positive when it should be negative... the derivative of that function is
\[\frac{-x^2+2x+8}{(x^2+8)^2}\]

Answer 4

yeah

Answer 5

so to get rid of the negative multiply by 1 (-1/-1)
\[\frac{x^2-2x-8}{-(x^2+8)^2}\]

Answer 6

wait you could do that

Answer 7

i'm not changing the equations by multiplying by one and -1/-1 is still 1

Answer 8

so instead of the bottom always being positive it will be negative

Answer 9

so evreything switches if you get a positive on the top it's negative if you get negative you get positive

Answer 10

so is -2 also relative maximum

Answer 11

well actually its probably relative minimum

Answer 12

-2 if you put it in

(x-4)(x+2)

test -3
you get (-7)(-1)=positive however the denominator is negative making it negative

test -1
you get (-5)(1)=-5 but the negative in the bottom cancels getting positive

since it was negative on the left side the function is decreasing and after it's increasing

-2 is a minimum

Answer 13

and yes you can multiply your equation by 1 at any time if you wish .. it sometimes can make something that was super ugly looking turn into something nicer looking (forms of 1 can be (x/x),(2/2)(1/1)(-1/-1)etc.

Answer 14

how come if I plug -3 into the original numerator which x^2-2x-8 I get a negative

Answer 15

then the bottom is negative which makes it positive

Answer 16

actually ignore that I did get a negative for the original numerator