VCE Methods Question Thread!

[Rachelle]

In the essentials exercise 2F, composite functions, a question reads:

f:{x:x≤3} --> R, f(x)=3-x and
g:R-->R, g(x)=x^2 - 1.

Show that f(g(x)) is not defined.
Define a restriction g* of g such that f(g(x))* is defined and find f(g(x))*.


How do I go about restricting g so that f of g can be defined?

Thankyou.

[wunderkind52]

In the essentials exercise 2F, composite functions, a question reads:

f:{x:x≤3} --> R, f(x)=3-x and
g:R-->R, g(x)=x^2 - 1.

Show that f(g(x)) is not defined.
Define a restriction g* of g such that f(g(x))* is defined and find f(g(x))*.


How do I go about restricting g so that f of g can be defined?

Thankyou.

For fog to be defined,

So, the range of g must be restricted to fall into this. Thus, solving for

So new domain of g is

[Rachelle]

For fog to be defined,

So, the range of g must be restricted to fall into this. Thus, solving for

So new domain of g is

Thank-you!

[Callum@1373]

Find the values of a and b for which x^3 + ax^2 -10x + b is divisible by x^2 + x - 12

also

let f(x) = x^3 - kx^2 + 2kx - k - 1
show that f(x) is divisible by (x-1)
factorise f(x)

thanks

[keltingmeith]

Find the values of a and b for which x^3 + ax^2 -10x + b is divisible by x^2 + x - 12

also

let f(x) = x^3 - kx^2 + 2kx - k - 1
show that f(x) is divisible by (x-1)
factorise f(x)

thanks

The factor theorem - if f(x) is divisible by (ax-b), then f(b/a)=0 So, for the first one:

If P(x)=x^3+ax^2-10x+b is divisible by x^2+x-12, it must also be divisible by the factors of x^2+x-12=(x-3)(x+4). So:

P(3)=0, 0=27+9a-30+b
P(-4)=0, -64+16a+40+b

We can simplify both to:

3=9a+b (1)
24=16a+b (2)

Now, we do (2)-(1) to get:
21=7a
a=3

Sub into (1):

3=27+b
b=-24

For the second, f(1)=1-k+2k-k-1=1-1+2k-k-k=0, so f(x) is divisible by x-1. You can then factorise using long division, or this:

f(x)=(x-1)Q(x), and we want to find Q(x). Q(x) must be a quadratic so that f(x) is a cubic, so we get:

f(x)=(x-1)(ax^2+bx+c)

Expanding:

f(x)=ax^3+bx^2+cx-ax^2-bx-c
x^3-kx^2+2kx-k-1=ax^3+bx^2-ax^2+cx-bx-c
x^3-kx^2+2kx-k-1=ax^3+(b-a)x^2+(c-b)x-c

Now, we "equate coefficients" - that is, we take the coefficients in front of specific pronumerals and equate them. So, for x^3:
1=a

For x^2:
-k=b-a

For x:
2k=c-b

For constants:
-k-1=-c

So, c=k+1 and a=1. Putting these into any of the above equations, we'll get that b=1-k. So:

f(x)=(x-1)(x^2+(1-k)x+k+1)

[Callum@1373]

Holy crap thanks man 

[KingDrogba]

Can anyone in the most basic and simple terms explain how to restrict the domain so that the function exists
In reference to question C of the attached

May Drogba be with you

[StupidProdigy]

Can anyone in the most basic and simple terms explain how to restrict the domain so that the function exists
In reference to question C of the attached

May Drogba be with you

Look two posts above your post..

[RazzMeTazz]

Can someone please explain how to do this question:

If -1/2 h(x+2) + 1 = -(x^3)/2 - 3(x^2) - 6x - 1/2. Find h(x) using matrix methods.

Thankyou

[keltingmeith]

Can someone please explain how to do this question:

If -1/2 h(x+2) + 1 = -(x^3)/2 - 3(x^2) - 6x - 1/2. Find h(x) using matrix methods.

Thankyou

Find the matrix that defines the transformations on h(x), then apply the inverse. (think of it like this: if you multiply h by 2, then to undo that transformation, you must divide by to - OR, apply the inverse of multiplying by 2)

[cosine]

How can i simplify:

[StupidProdigy]

How can i simplify:

Rationalise the denominator?

[Sine]

How can i simplify:

Multiply by it's conjugate:   

[wobblywobbly]

Multiply by it's conjugate:

So it seems we have a sine and a cosine....

[cosine]

Rationalise the denominator?

How though? If you multiply by it's conjugate, then why is it that     you multiply by root 2, and not the conjugate?

So it seems we have a sine and a cosine....

It seems so... -.-