IGCSE MATHS Doubts

[Vin]

18 b)

(x^-2)^-1/2
------
(2²)^-1/2

x¹
----
2^-1

= 2x

[-Cornea]

i)(x^27/27)^2/3
=(x^27*2/3)/(27^2/3)
=(x^9*2)(divide by 3) (cube root of 27 =3 square it =9)
so it will be (x^18/9)
as27^2/3 means cubic root and then squaring it

[Malak]

Thx

[Malak]

07 Jun P2 Q16 Part a

f(x) = 3x - 1

Find. in its simplest form

F^-1f(x)

In ms the answer is x how

[astarmathsandphysics]

x
y=3x-1 so x=(y+1)/3 so f^(-1)( f(x))=(f(x)+1)/3=(3x-1+1)/3=x

[Vin]

f-1(x) =       x + 1
                   3
                         
put 3x - 1 as x in eqn 1 ..u'll get it

(3x -1) - 1
-----------
     3


= x

[SEL123]

I get really confused when we're asked questions on shear and stretch..

I asked my tutor and he said for IGs the invariant like is ALWAYS either the x-axis or the y-axis...

To someone who has practiced past papers

Is that true??

[Fastal]

listen when u do f(x) u remove the X and replace it by the number given
now it's f^-1(x) so replace the y by 1 instead of replacing x because it's inverse
so it will be:
1=10^x
Anything to the power of zero =1
so x =0

listen u just need to find the perimeter of the shape
4*4 i.e(4+4+4+4) +2*pie*1.2 i.e(4*0.25*2*pie*1.2)
i guess the answer will be 23.5 approximatly it's 24 ppl (or if u want to add this half person )

i dont understand the function part.



and i still dont get how its vertically left for that shear question in 09November paper 4

[$H00t!N& $t@r]

i dont understand the function part.

The function given to you is:   f(x)= 10^x
The question says: write down the value of f^-1(1)
Take the 1 in the bracket from f^-1(1) and put it on the other side
So now it becomes 1= 10^x
One of the law in indices is the zero index law which says that any number raised to the power of 0 is equal to 1
After rearranging we have 1 as the answer so that means that 10 is raised to 0
Therefore x=0
Hope that helps!  

[Fastal]

Take the 1 in the bracket from f^-1(1) and put it on the other side
So now it becomes 1= 10^x

i dont understand this part, i get everything else

[-Cornea]

inverse the normal things u do:D tht's it

[$H00t!N& $t@r]

i dont understand this part, i get everything else

Look, in the question before this (7a) the question was f(0.5) so u replace the x in f(x)=10^x with o.5 and you get 3.16

However, this is not the case in the second one. Here you're meant to find the inverse f^-1 (1)
so it will be the other way round. The 1 in the bracket (f^-1 (1)) will be replaced with f(x)  in f(x)=10^x
giving you 1= 10^x

If you still dont understand can you please specify what is it that you dont understand

[-Cornea]

may june 2008 construction and locus question plzzz

[Fastal]

Look, in the question before this (7a) the question was f(0.5) so u replace the x in f(x)=10^x with o.5 and you get 3.16

However, this is not the case in the second one. Here you're meant to find the inverse f^-1 (1)
so it will be the other way round. The 1 in the bracket (f^-1 (1)) will be replaced with f(x)  in f(x)=10^x
giving you 1= 10^x

If you still dont understand can you please specify what is it that you dont understand

hmm, we have a different way of finding inverse in school
suppose f(x)=2x+1
than f^-1(x) is replace by y

y=2x+1(y and x are then switched)

x=2y+1
thus y=(x-1)/2
and f^-1(x)=(x-1)/2

so can you tell me how to do it this way what is F^-1(x) when f(x)=10^x

[-Cornea]

that's right inverse it in this way:
but if it's like this question instead of removing x remove y
listen f(x)---> remove x
        f^-1--->opposite remove (y)
Got it:D