Open page test- (1 Viewer)

[shinji]

hey guys, i've got an open page test coming up on the subjects of Inverse Fuctions and Inverse trig functions and also Integration.

I can only write one a4 page to bring into the exam, so what should i put in?

i'm currently thinking of putting in:
- finding inverse functions
- domains and ranges of inverse functions (that they're swapped around)
- General Solutions
- Transformation Method (aka R Method)
- Summary of the inverse trig functions: Sin^-1, Cos^-1, Tan^-1
- Quetsions like Sin(Cos^-1t/₄
- Prooving inverse trig differentials

- Methods of integration Sin²x, Cos², etc.
- Other Methods of Integration (Cos³Sinx

and that's all so far. i don't have much on methods of integration as it is particularly a small topic.


here are also some practise questions that my teacher gave us that i have trouble with ..
1) By restricting the domain of f(x) = x² = x² + 6x, find the inverse function over that domain.
im pretty sure that that's easy .. but im stuck on it. <_<"

2)State the domain and range of y = sin^-1[tanx]

3) Differentiate xcos^-1x - [sqrt]1-x²[/sqrt]

4) find the area bounded by the curve y = cos^-1x, the x-axis and the lines X=0 and X=1/2


Thanks a lot in advanced!

 

[pLuvia]

1)
Restrict the domain either x>-3 or x<-3
Then just find the inverse function by swapping the x and y

2)
Domain of sin^-1x is -1<x<1
Domain of sin^-1(tanx) is tan(-1)<tanx<tan(1)

Range of arcsinx is -pi/2<y[]<[/u]pi/2
Range of arcsin(tanx) is the same

3)
d/dx=xarccosx - sqrt{1-x²}
=arccosx+x/sqrt{1-x²}+x(1-x²)^-1/2

4)
I can only think of this method right now
"Integration by parts"

But I think you can use a rectangle method to get the answer

With regards to your test
I think all that should be sufficient

 

[Riviet]

A couple more useful formulae that you might want to add to your summary sheet:

sin^-1x + cos^-1x = pi/2

cos^-1(-x) = pi - cos^-1x

 

[pLuvia]

A couple more useful formulae that you might want to add to your summary sheet:

sin^-1x + cos^-1x = pi/2

cos^-1(-x) = pi - cos^-1x

That might be one of the questions in your test that you have to prove shinji

 

[shinji]

hehe. i know;
sin-1x + cos-1x = pi/2
i know how to prove... but cos^-1x = pi/2 ??

hehe, thanks for ur help though!

 

[SoulSearcher]

hehe. i know;
sin-1x + cos-1x = pi/2
i know how to prove... but cos^-1x = pi/2 ??

hehe, thanks for ur help though!

Easily proven:
Let @ = cos^-1(-x)
Then -x = cos @, where 0 < @ < pi
So cos(pi-@) = x, since cos(pi-@) = -cos @
pi - @ = cos^-1x
@ = pi - cos^-1x
Therefore
cos^-1(-x) = pi - cos^-1x

 

[shinji]

3)
d/dx=xarccosx - sqrt{1-x2}
=arccosx+x/sqrt{1-x2}+x(1-x2)^-1/2

for that, i keep on getting just cos^-1x

coz for the first part, u'd use the product rule. would would =
y'= arcCosX + x/sqrt{1-x²} - x/sqrt{1-x²}
= arcCosX .. as they latter 2 cancel out. Note; i got the negative by the chain rule .... or whatever one it is. where u differentiate in the bracket and u times it by the power.

and then would u use the solution of this differential for the integral of the #4?

IRT Soul-Searcher: thanks for that! . makes sense now.. lol

 

[SoulSearcher]

d/dx (xarccosx - rt(1-x²) = arccosx - x/rt(1-x²) + x/rt(1-x²)
=arccos x
Your answer should be right, the same question appears in Cambridge, and their answer is arccos x