Methods Q's

[sxcalexc]

I have no idea how this question can be solved... Appreciate any help



Find the coordinates of the point of intersection of the graphs of and . Express the coordinates correct to 2 decimal places.

This has to be a calculator question, right? Well, the only thing I can do on the calculator is draw the inverse. Is there any way to calculate the actual equation and/or calculate the intersects.


As for the second question I attempted it and I'm not certain whether I got it right.

Find if has a domain of and give the domain of

I transposed the equation to get . I then used the quadratic formula and swapped the y's and x's.
I ended up with
I tried graphing it and it isn't looking the same as the drawn inverse.

Thanks in advanced.

[ed_saifa]

For the first question you must let f(x)=x for the intercepts with the inverse.Since the function is reflected in the line y=x.
For the second question, the domain and ranges are swapped with the inverse. So, the domain of f(x) is the range of its inverse. You need to reject one of the solutions as there is a restriction.

[sxcalexc]

For the first question you must let f(x)=x for the intercepts with the inverse.Since the function is reflected in the line y=x.
For the second question, the domain and ranges are swapped with the inverse. So, the domain of f(x) is the range of its inverse. You need to reject one of the solutions as there is a restriction.

Thanks for your fast reply. With the first question, when I make the equation = x I only get 1 intercept. However, when drawing the inverse it appears there are more intercepts within the domain.

With the second question, both equations (+-) fit within the range :S
Here is a computerised graph of the functions..
The black line was hand drawn in paint and is a rough drawing of the inverse the calculator produces
The red line is the + equation, the green line is the - equation

[Mao]

the first question would have to be a calculator question, yes.

The inverse doesnt necessarily have to be when . theres a post about it *go digs it up*

solving in y=x is generally accepted, but sometimes there can be multiple (or infinite) solutions, such as a symmetry of y=x already exists:

(a quater circle)
which the reverse is itself... (infinite solutions)

what you need to do is with your solver on the calculator, find when and be prepared to wait a minute or so

I doubt an exam question will be this difficult,
are you doing CAS?


as for the second question, you are thinking it a lot more than what it is:

Range of F becomes Domain of F^-1
Domain of F becomes Range of F^-1

its that simple
all you need to find is the minimum and maximum values of

[sxcalexc]

the first question would have to be a calculator question, yes.

The inverse doesnt necessarily have to be when . theres a post about it *go digs it up*

solving in y=x is generally accepted, but sometimes there can be multiple (or infinite) solutions, such as a symmetry of y=x already exists:

(a quater circle)
which the reverse is itself... (infinite solutions)

what you need to do is with your solver on the calculator, find when and be prepared to wait a minute or so

I doubt an exam question will be this difficult,
are you doing CAS?


as for the second question, you are thinking it a lot more than what it is:

Range of F becomes Domain of F^-1
Domain of F becomes Range of F^-1

its that simple
all you need to find is the minimum and maximum values of

Thanks for the help nah, I'm doing normal methods.

It appears the answer was slack for the first question. They only made it = x and found 1 point of intersection when there are actually several

For the second question it did ask for the inverse. I got it half right ~ it only wanted the positive solution because y > 0, which i don't get because the negative solution is in y > 0 also.

[Mao]

righto, cool

the thing about inverse function is:
the function HAS to be one-to-one

graphically, it is a function inverted in y=x. that means many-to-one functions (the functions that fail horizontal line test) dont have an inverse function, they have an inverse relation, which may or may not use the variable y implicitly.

in some cases, the domain need to be restricted so that the function is one-to-one, and an inverse for it can be found.

but in this case, is not one-to-one for ...
its a wierd question
and as for "the negative solution is in y > 0 also", are you sure? for all x<0, y<0
it is a wierd question

[sxcalexc]

Ah, so the whole function must be y>0 ? but the domain is positive, so shouldn't it apply ? Haha im confusing myself

[Mao]

the function does not need to be within a certain range (even though it might be specified)

what need to happen for a function to have an inverse is it must be one-to-one
i.e. pass the horizontal line test.

such as for , it doesnt have an inverse as you can draw a horizontal line that passes through more than one points. (the inverse relation here will be , which you might write as )
if, however, we were to restrict it to , it will have an inverse function.

this question, however, is stupid, as it doesnt pass the horizontal rule test for , hence you'll end up with an inverse relation like you have found.
the proper restraint should be
but this doesnt stop you from working out the range of for , which is , and that will be the domain of your inverse relation

[sxcalexc]

Sweeet thx Mao