My Methods 3/4 questions thread

[brightsky]

you do not need to use a matrix to determine the inverse of a particular function. simply swap x and y and solve for y. however, if you prefer the matrix method, then the transformation matrix is [0,1;1,0]. this reflects the original graph in the line y=x, as you would do to find the graph of the inverse.

[Einstein]

how do we divide polynomials by inspection?

[Phenomenol]

how do we divide polynomials by inspection?

(x - 1) and (x + 1) are the first ones you should try.

Also try looking at the constant (the x^0 coefficient). For the polynomial to factorise into the form (x - a)(x - b)(x - c)... where a, b, c.. are integers, then a, b and c etc. will all be factors of the constant.

Sometimes the factors are easy to see in the first place. For example, x^3 + 2x^2 + x + 2 = x^2(x + 2) + (x + 2) = (x^2 + 1)(x + 2).

[smile+energy]

you do not need to use a matrix to determine the inverse of a particular function. simply swap x and y and solve for y. however, if you prefer the matrix method, then the transformation matrix is [0,1;1,0]. this reflects the original graph in the line y=x, as you would do to find the graph of the inverse.

Awesome

[smile+energy]

Hi, 

can I get some help?
I don't understand what the question is asking: for the following function, fully define its inverse
f:[-2, infinite)->R, f(x)=(x-2)^2-3

[Phenomenol]

Hi, 

can I get some help?
I don't understand what the question is asking: for the following function, fully define its inverse
f:[-2, infinite)->R, f(x)=(x-2)^2-3

To fully define the inverse function, you need an equation and a domain.

[IndefatigableLover]

Hi, 

can I get some help?
I don't understand what the question is asking: for the following function, fully define its inverse
f:[-2, infinite)->R, f(x)=(x-2)^2-3

Well with the function you have now, it doesn't have an inverse function (only one to one) but even with the restricted domain, you still don't have an inverse function (though I assume it to be a typo since '-2' should be '2').

I'm sure you know how to work out the inverse of that equation

Spoiler

Answer is: since they want the positive side of it rather than the other side

but you need to worry about the domain/range. In this case you know that the range of f(x) will be the domain of the inverse so if you sub '2' into f(x) then you'll find that your range for f(x) = [-3, infinite).

Therefore you can write up your final answer as:

[smile+energy]

Thanks a lot, guys. 

[Einstein]

is functions hard in units 1/2, were upto cubics and functions looks intimidating, all those symbols as heaps of f: haha

[rhinwarr]

Don't worry, mostly you'll just be learning the shapes of the graphs because once you learn transformations for one type of graph, they are applied in the same way for pretty much all the other functions. (eg. co-efficient of x always dilates the graph from y axis etc.)

[knightrider]

how do you draw a function when there is an unknown factor?

For example f(x)=k(x-2)^2+3

[lzxnl]

You can't really

[Phenomenol]

how do you draw a function when there is an unknown factor?

For example f(x)=k(x-2)^2+3

Depends. In your example, the y intercept would be in terms of k, but the turning point won't, and there will be no x intercepts.

[lzxnl]

Depends. In your example, the y intercept would be in terms of k, but the turning point won't, and there will be no x intercepts.

You actually can't sketch that because you don't know if k is positive or not. A positive k will have no x-intercepts, but a negative k will. k thus determines the shape and the behaviour of the graph

[Phenomenol]

You actually can't sketch that because you don't know if k is positive or not. A positive k will have no x-intercepts, but a negative k will. k thus determines the shape and the behaviour of the graph

Ah yes. My mistake.