Question

f(x)= (x-2)(x+4)
I need help finding the intervals that its increasing and decreasing and find critical point

Answer 1

so to find my critical point I take the derivative and equal it to zero right?

Answer 2

Yes

Answer 3

and to check my local min and max I plug my critical number into the original function?

Answer 4

Yes.

Answer 5

ok. I got -1 for the critical number and I plugged it back in the original function but I got -9...

Answer 6

i also did the first derivative test and found out that my local min was -1. and that my local max was 1. so i tried finding concavities by getting the second derivative and making that equal zero but here was my problem

Answer 7

my second derivative became just 2. a constant. I have no x's in my equation. did I do something wrong? shouldn't I have an x in my second derivative so that I can have something to plug my x values in so that I can see where I have my concavities?

Answer 8

does this mean I have no concatives?

Answer 9

I hope that all makes sense. Sorry for the overload. I'm kinda confused.

Answer 10

i really need help!

Answer 11

anyone?

Answer 12

f(x) = x^2+2 x-8

Answer 13

i tink

Answer 14

ok???

Answer 15

You are doing well, I think you understand it. Also you are the only actual person who asked a legitimate question in the history of OpenStudy.

Answer 16

Because of that I know someone who will explain it better than me

Answer 17

@SithsAndGiggles Someone asked for legitimate help could you assist, Thank you!

Answer 18

@.Sam.

Answer 19

thanks but I really need this question answered

Answer 20

:(

Answer 21

There is no concavity is why : )

Answer 22

@sohrinny

Answer 23

Ooops, @SithsAndGiggles to the rescue.

Answer 24

\(f(x)\) is just a parabola, so it's either concave up or down for all \(x\). There are no inflection points.

Answer 25

@sohrinny, like you mentioned, \(f''(x)=2\), and to find inflection points you solve for \(f''(x)=0\).

Well, now you have the equation \(2=0\), which isn't true. So there are no solutions for \(x\), i.e. no inflection points.

Answer 26

ok thanks!

Answer 27

but my steps weren't wrong?

Answer 28

So how did we get a local max of 1? Were you given an interval or just the function?
@sohrinny