Question

Pre-Calculus question.
Increasing, decreasing; local minimum/maximum.
Not sure whether to use derivative or not as I'm 87% sure I haven't learned that (isn't that Calculus anyway?)

http://prntscr.com/a2y6z8

Answer 1

Yeah it's Calculus, but can't you just use graphing calculator?

Answer 2

I tried but I got stuck > <;
(also I haven't slept since 1 pm yesterday and it's 4 pm so o_o)

Answer 3

http://www.algebra.com/algebra/homework/Functions/Functions.faq.question.198715.html this one says use derivative so > >
I'm not sure. I'm struggling with my graphing calculator

Answer 4

Okay here's what I got off Wolfram http://prntscr.com/a2ybxy

Answer 5

Calculus is an easy way to solve, but since you haven't learn, graphing is your best shot.

Here's I have graphed it for you: https://www.desmos.com/calculator/ant7xs7pl3

In between local minimum/maximum or discontinuity, it either increases or decreases, right? You should be able to tell from graph

Answer 6

OH I FORGOT ABOUT THAT WEBSITE > <

Answer 7

So there are four intervals: \((-\infty, -\sqrt{21}), (-\sqrt{21}, 0), (0, \sqrt{21}), \)and \((\sqrt{21}, \infty)\)

Answer 8

Intervals in middle should be easy part. YOu got this?

Answer 9

It wants decimal form, so it's similar to the min and max I have earlier in the problem, yes?

Answer 10

http://prntscr.com/a2yei8

Okay I got the intervals somewhat

Answer 11

Yeah, \(\sqrt{21}\approx4.58\)

Answer 12

do I have to test the intervals? o-o

Answer 13

Nah, if you "read" graph from left and right, if it goes down, it decreases, but if it goes up, then it increase. That makes sense, right?

Answer 14

Yeah but I thought the asymptote was\[x=\sqrt{21}\]

Answer 15

Also I don't see where it goes up,that's the issue > <

Answer 16

Asymptote is only \(x=0\)

But you can see where it goes down, right?

Answer 17

Yes

Answer 18

Okay, right after \(\sqrt{21}\), does it increase or decrease? you can zoom in to see closely lol

Answer 19

Okay @ @

Answer 20

It seems to flatline a bit then dip down slowly

Answer 21

I compressed y-axis for you; it should be easy now

https://www.desmos.com/calculator/2xd6cosy27

Answer 22

OH, I see. That was on Wolfram's secondary graph as well

Answer 23

It dips pretty fast

Answer 24

You can change axes range in graph settings on top-right :)

Answer 25

Okay :]

Answer 26

dip? that's another word for decrease?

Answer 27

because it looks increasing to me

Answer 28

At \(x=\sqrt{21}\) and forward, it increases

Answer 29

Oh yeah, it increases then drops down fast* sorry lol

Answer 30

Oh well that's for \(x<0\), increase slowly then go down fast.

Opposite goes for \(x>0\); drop down fast then increase slowly

Answer 31

Q: When do I know to adjust the y-increments?

Answer 32

So now you know which intervals are increasing and decreasing, right?

Answer 33

you do that when it looks flat, you change y-increments to "zoom in"

Answer 34

Er... still a little confused. I get the increasing part but where does it stop increasing o-o

Answer 35

for \(x>0\), it never stops lol

Answer 36

it just increase slower and slower and slower...

Answer 37

It is approaching to 0. but it wont reach 0.

Answer 38

So then...
http://prntscr.com/a2ykdt

Answer 39

You do agree that \((-\sqrt{21}, 0)\) and \((0, \sqrt{21})\) are decreasing intervals, right?

Answer 40

Yes, I believe they are. At least from a left-to-right perspective.

Answer 41

Yeah, so put that for decreasing.

Then put the remaining intervals for increasing part.

Answer 42

Wait what x x;

Answer 43

There are four intervals: \((-\infty, -\sqrt{21}), (-\sqrt{21}, 0), (0, \sqrt{21}), \)and \((\sqrt{21}, \infty)\)

\((-\sqrt{21}, 0), (0, \sqrt{21})\) are decreasing

\((-\infty, -\sqrt{21}), (\sqrt{21}, \infty)\) are increasing

Answer 44

Sorry for confusion!

Answer 45

okay ascending: (-inf, -4.58) U (4.58, inf)
descending: (-4.58, 0) U (0,4.58)??

Answer 46

Yesss I lagged sorry > <

Answer 47

But thank you ^ - ^

Answer 48

Looks good

Answer 49

Did I help clearing thing up?

Answer 50

Yes :D Thank you very much!! ☺

Answer 51

No problem

Answer 52

Oh yes I might also need help on this one - http://prntscr.com/a2yng7 @geerky42 (if you're busy then it's okay! ^^; )

Answer 53

25 mph is correct :)