Mr. Study's 3 and 4 Questions

[Mr. Study]

Hey,

Thought I would start this topic, so I can tackle my problems with Methods!

Essential Math Methods

Question 1: f(x)=3x^2+x-2. Find {x:f(x)>0}
Could someone explain what this means: {x:f(x)>0} and then show me how to solve it?

Question 2:
I can transpose other log equations to find x but this one has me stumped, as it's not the usual equation for me. (If that makes sense.  )

Question 3:

Question 4:
Its suppose to be f circle g and not the word fog. I have never seen the circle symbol used before in math. xD

Question 5:

If someone explains Question 1 dont worry about Question 5.

Thanks!!

[Greatness]

1) It means find the values of x for which the function is greater than 0. There are probably a few ways to do this, you can tell its a positive parabola so the values of x which are positive are the left side of the left x intercept and the right side of the right intercept - if that makes sense to you. lol
2) x= 10^2  because log b (a) = p so a=b^p
3) skip lol
4) i think it's a composite function... lol really cant remember
5) its asking for when the derivative is greater than 0

[Mr. Study]

1. Ahh!! I get it! I was thinking of some other things and NOT of it being positive.
2. Ah, I was thinking of doing it like that but I wasn't too sure it was the correct way of showing working out.

3. xD

4. Out of curiousity, What is a composite function?
5. So would I do f'(x)=6(x-1).
    Then let f'(x) = 0
    So it looks like 0 < 6(x-1)
                           0  < x-1
                            1 < x
Am I on the right track?? I didn't do enough of these types of problems last year.

Just incase. No need for Questions 1 and 2 now. Yay!

Whoops. Thank you for the help swarley.

[Greatness]

4. http://en.wikipedia.org/wiki/Function_composition you'll learn it some stage of the year, i hated them.
5. looks good

[Mr. Study]

5. It does??? Wow.

Thank you so much for your help swarley.

[Greatness]

5. It does??? Wow.

Thank you so much for your help swarley.

Yeah i think so.... Altho i havent touched any math since exams... Hopefully most of what i've said is right

[Mr. Study]

Heres my question:

I cant really draw the cubic graph on paint/microsoft but heres all the information.

Maximum turning point: (-1,6)
Minimum T.P: (1,4)
Y-Intercept: (0,-1)

Find {x:f'(x)>0}. We are also not given the cubics equation.

I checked the answers but it is {x:x<-1} u {x:x>1}. I am not too familiar with that way of writing so could someone also explain what it essentially means?

Thanks! 

[pi]

Removed last post, just realised my mistake (a big one!), will repost soon


edit: I give up, someone else find an equation to the cubic. For me, the y-int does not add up I think I've forgotten all of methods lol! Ignoring the y-int, the equation can be given by: ... I think. But that doesn't add up to the info, sorry Mr. Study

[Phy124]

Heres my question:

I cant really draw the cubic graph on paint/microsoft but heres all the information.

Maximum turning point: (-1,6)
Minimum T.P: (1,4)
Y-Intercept: (0,-1)

Find {x:f'(x)>0}. We are also not given the cubics equation.

I checked the answers but it is {x:x<-1} u {x:x>1}. I am not too familiar with that way of writing so could someone also explain what it essentially means?

Thanks! 

Given those turning points, I don't think the y-intercept can be -1? (I'm with you Rohit )

Let's pretend it says the y-intercept = (0,5)

We want to find when f'(x) >0 (When the graph is increasing)

Given the stationary points and our intercept, our graph will look like this https://www.desmos.com/calculator/kb0hymstlt (The equation Rohitpi worked out)

Therefore, it is increasing from (-oo,-1) then decreasing (-1,1) and increasing again from (1,oo)

We only want when it is increasing so

(-oo,-1) and (1,oo)

Which can also be written as

{x:x<-1} and {x:x>1}

If anyone else can find a logical solution that doesn't involve changing what is given then by all means call me out

[b^3]

Agreed that the y-intercept is pretty screwed up. Also to add to Phy124's graph
https://www.desmos.com/calculator/nsivedqhna
f'(x)>0 when that tangent has a positive gradient, i.e. when it is not between those two green lines, try moving the slider for the value of a and having a look at what happens

[Mr. Study]

Heres the image, I probably should've uploaded it but I couldn't find mine CD.
.

Thank you all for the help, I actually get it now!!

[b^3]

Heres the image, I probably should've uploaded it but I couldn't find mine CD.
(Image removed from quote.).

Thank you all for the help, I actually get it now!!

Makes sense then, (1,-4) not (1,4). But yeh in this case, you can see by the graph when the tangents will be positive, without having to find an equation.

[Mr. Study]

Ah crap, Sorry about that!!!

[Mr. Study]

Todays theme is Log questions. (Woo... ).

.

I'll try Latex next time. (Hopefully. ).

Thanks

[pi]

Probably made some stupid mistakes, but this is the way to go about them


1.





2.
Let


or
or
or

3.
Multiply both sides by


or
as