VCE Methods Question Thread!

[Zues]

hey euler, thanks!

say i had the equation y = a sqrt (bx+c) + d, a=/0

the endpoint would be (-c/b , d). but if it asked for the dilation from the y axis it would be 1/b, and from the x axis it would be a*sqrt b ?

also, looking at earlier posts i still dont entirely get why the y axis dilation is 1/n, could you quickly explain it in a easily comprehendible manner. I know how to apply it, but might be interesting to know why its that.

thanks

[Zues]

Q
Sketch the graph of the following functions and state the domain and range of each.
f(x) = sqrt (4 - x^2)

hence, find the inverse of the function . Sketch the graph and state the domain and range of each inverse.

swapping y an x in the equation and solving for y produces the same equation? how do you even go about graphing f(x) = sqrt (4 - x^2) as well?

[pi]

Well this is obviously a semi-circle of radius 2 and centre (0,0) Might help to rearrange to a form of x^2 + y^2 = r^2 to see this.

y = sqrt(4 - x^2)

Swapping x and y to find inverse

x = sqrt(4 - y^2)

x^2 = 4 - y^2

y = +/- sqrt(4 - x^2)

Therefore f-1(x) = +/- sqrt(4 - x^2)

This is NOT the same as the original equation.

However, recalling that f(x) was not a one-to-one function to begin with, it needs to be restricted in its domain, say [0,2] to create a one-to-one function. So we'll take the positive f-1(x) = sqrt(4-x^2) accordingly with a domain restriction of [0,2] (which was the range of f(x)) and you're done!

[SE_JM]

Hello, I need a little confirmation
There is this question:
Find the inverse function of x² and x⁴

I said the inverse function is + and -x^1/2 and + and -x^1/4 respectively

But the correct answer tells me that the answer is +x^1/2 and +x^1/4 only
Am I right? I think so... But maybe not

[pi]

Hello, I need a little confirmation
There is this question:
Find the inverse function of x² and x⁴

I said the inverse function is + and -x^1/2 and + and -x^1/4 respectively

But the correct answer tells me that the answer is +x^1/2 and +x^1/4 only
Am I right? I think so... But maybe not

I don't remember this well, but y=x^4 is not a one-to-one function, so you have to split it up. Say we take the largest possible domain so that f-1 exists, we'll take [0,inf). If we inverse the function with that restriction in mind we'll only take the positive x^(1/4) with a domain of [0,inf) (this is the range of the restricted function!).

[SE_JM]

I thought x⁴ was one-to one function:
http://www.wolframalpha.com/input/?i=x%5E4

[Zues]

Well this is obviously a semi-circle of radius 2 and centre (0,0) Might help to rearrange to a form of x^2 + y^2 = r^2 to see this.

y = sqrt(4 - x^2)

Swapping x and y to find inverse

x = sqrt(4 - y^2)

x^2 = 4 - y^2

y = +/- sqrt(4 - x^2)

Therefore f-1(x) = +/- sqrt(4 - x^2)

This is NOT the same as the original equation.

However, recalling that f(x) was not a one-to-one function to begin with, it needs to be restricted in its domain, say [0,2] to create a one-to-one function. So we'll take the positive f-1(x) = sqrt(4-x^2) accordingly with a domain restriction of [0,2] (which was the range of f(x)) and you're done!

mhm i havent learned about semi circles. i would assume the following

y = sqrt(4 - x^2)
y^2 = 4 - x^2
y^2 + x^2 = 4, now isnt this a circle? with centre 0,0 and radius 2?

then i wouldve swapped x and y hence
x^2+y^2 = 4, same thing?

with your last sentence
However, recalling that f(x) was not a one-to-one function to begin with, it needs to be restricted in its domain, say [0,2] to create a one-to-one function. So we'll take the positive f-1(x) = sqrt(4-x^2) accordingly with a domain restriction of [0,2] (which was the range of f(x)) and you're done!

if f(x) was not a one to one, why does the following inverse also need to be restricted? i know the range of f^-1(x) will take the domain of f(x) but i havent come across needing to restrict  domain. And where did the restriction of the domain [0,2] come from?

thanks

[IndefatigableLover]

I thought x⁴ was one-to one function:
http://www.wolframalpha.com/input/?i=x%5E4

It's not a one-to-one function since x^4 doesn't pass the vertical/horizontal line test. Similarly to x^2, it is a many-to-one function since when you use the horizontal line test, it cuts the graph twice.

[pi]

Zues, only one to one functions have inverses, any restriction you make to a function to ensure it is one to one is passed onto its inverse too.

[Zues]

It's not a one-to-one function since x^4 doesn't pass the vertical/horizontal line test. Similarly to x^2, it is a many-to-one function since when you use the horizontal line test, it cuts the graph twice.

i thought to be a function it just needs to ONLY pass the vertical line test? i can understand why it needs to pass the horizontal otherwise its a many to one if it hits twice, but this is what my book says

A one-to-one function is a function where for each x-value there is only one y-value and vice versa. The graph of a one-to-one function can be crossed only once by any vertical or horizontal line.
A function that is not one-to-one is many-to-one.... but then it says this?
A function will have an inverse that is also a function if and only if it is a one-to-one function. If a
function, f , is one-to-one, then its inverse function is denoted by f −1.

[Zues]

Zues, only one to one functions have inverses, any restriction you make to a function to ensure it is one to one is passed onto its inverse too.

so lets say find the inverse of f(x) = x^2 + 5x + 3

they say it can only be one to one to have an inverse. however this produces an inverse of f^-1(x) = +  sqrt (x+13/4) - 5/2  OR   f^-1(x) = - sqrt (x+13/4) - 5/2

[SE_JM]

oops. Got confused with it being a function or not. sorry

"Say we take the largest possible domain so that f-1 exists, we'll take [0,inf). If we inverse the function with that restriction in mind we'll only take the positive x^(1/4) with a domain of [0,inf) (this is the range of the restricted function!)."

does this mean, I can take (-inf, 0] instead and take -x^(1/4) with domain [0, inf) as well?

Thank you!

[Zues]

i see that e.g. x^2 is a function but its a many to one, hence no inverse? but if i draw a y = x line it can be found to have an inverse? im getting a bit confused haha...

with inverses, it says if its not a one to one or manyto one its inverse does not exist. however in earlier chapters they were doing inverses of circles, quadratics, cubics etc.
e.g. f (x) = 5 − x^2, f(x) is a parabola reflected in the x-axis and translated 5 units up. It is not a one-to one function and so f −1(x) does not exist.

however you can do an inverse?, i know the output MAY bring about a +- x.... but then couldn't you say f^-1(x) = +.... OR f^-1(x) = -....

[SE_JM]

i thought to be a function it just needs to ONLY pass the vertical line test? i can understand why it needs to pass the horizontal otherwise its a many to one if it hits twice, but this is what my book says

A one-to-one function is a function where for each x-value there is only one y-value and vice versa. The graph of a one-to-one function can be crossed only once by any vertical or horizontal line.
A function that is not one-to-one is many-to-one.... but then it says this?
A function will have an inverse that is also a function if and only if it is a one-to-one function. If a
function, f , is one-to-one, then its inverse function is denoted by f −1.

Wait, my teacher told me the same
I'm kinda confused now.

So, a function can only have an inverse if it's a one-to- one function only?
Because i thought any graph given that it's a function could have an inverse..

[pi]

i thought to be a function it just needs to ONLY pass the vertical line test? i can understand why it needs to pass the horizontal otherwise its a many to one if it hits twice, but this is what my book says

A one-to-one function is a function where for each x-value there is only one y-value and vice versa. The graph of a one-to-one function can be crossed only once by any vertical or horizontal line.
A function that is not one-to-one is many-to-one.... but then it says this?
A function will have an inverse that is also a function if and only if it is a one-to-one function. If a
function, f , is one-to-one, then its inverse function is denoted by f −1.

As far as Methods goes:
- Relation: anything under the sun as far as the course goes
- Functions: vertical line test (ie. one-to-one and many-to-one)
- Functions with inverses: vertical line test AND horizontal line test (ie. one-to one)

Many-to-one functions (such as the semi-circle you provided) need to be restricted in its domain to be made into a one-to-one function if you want to find an inverse FUNCTION.

oops. Got confused with it being a function or not. sorry

"Say we take the largest possible domain so that f-1 exists, we'll take [0,inf). If we inverse the function with that restriction in mind we'll only take the positive x^(1/4) with a domain of [0,inf) (this is the range of the restricted function!)."

does this mean, I can take (-inf, 0] instead and take -x^(1/4) with domain [0, inf) as well?

Thank you!

Again I'm not well versed with that having not done maths for about 3 years haha, but theoretically, yes. However, they pretty much always assume the largest domain or state that in the question.

Maybe someone who has done the course more recently like Euler, IL, etc can confirm.