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#### Justanotherhuman

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##### Specific methods questions
« on: April 06, 2019, 05:19:00 pm »
0
Hey guys,
I have a few methods questions - if you could explain it to me it's be great!

1: How does
1/x + 3 = 1/x-3
= 3x^2 - 9x -3 = 0
(I think I lost track of how we get there)

2: In finding the maximal domain and range for root(16-x^2)
16 - x^2 >= 0
-x^2 >= -16
x^2<= 16
x<= +4 / -4
That indicates that x is less that or equal to +4 and +4
Where did I go wrong?

3: Implied domain for root(x^2 -x -2)
(x-2)(x+1) cannot equal 0
x cannot equal 2
x cannot equal -1
however the answer is (-infinity, -2) U (1, infinity)
Where did I go wrong?

Again, your help is very appreciated

#### colline

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##### Re: Specific methods questions
« Reply #1 on: April 06, 2019, 06:01:04 pm »
+3
Hey there! I'm a current student as well but I'll give your questions a go:

For the second question:
Try drawing the graph of f(x)=16-x^2 first.
Then, see what part of the graph is bigger than or equal to zero (i.e. above the x-axis).
With inequalities, it's always best to think about it graphically, so that you can actually SEE what is going on.

For 3, remember that when you square root a number, the number can never be NEGATIVE.
Then you draw the graph of f(x)=(x-2)(x+1) to see what part of the graph lies above the x-axis.
It's dividing when the domain excludes zero.

I'm not too sure about the first question. Are you sure you got the original equation right? It doesn't seem to work.

Hope the second and third questions make sense now
« Last Edit: April 06, 2019, 06:05:20 pm by colline »
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#### Justanotherhuman

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##### Re: Specific methods questions
« Reply #2 on: April 06, 2019, 06:29:10 pm »
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Hey there!
Wait so does that mean inequalities are not always right lol
But the 1st equation is right
The original question was to find the inverse function of f(x)= 1/x  + 3 and find point of intersection.
The inverse is 1/x-3
So the working out says
(1/x) +  3 = 1/x-3
and goes straight onto 3x^2 - 9x - 3 = 0

How would you find the implied domain of x-1/x+2? How would you sketch it?
« Last Edit: April 06, 2019, 06:37:23 pm by Justanotherhuman »

#### Justanotherhuman

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##### Re: Specific methods questions
« Reply #3 on: April 06, 2019, 06:32:44 pm »
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Also, did you just memorised the general shape and equation of every graph? I find it hard to draw a graph given an equation atm

#### colline

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##### Re: Specific methods questions
« Reply #4 on: April 06, 2019, 11:24:59 pm »
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Hey! Sorry for my late response!

For Q1, can you please specify whether the question is asking for the inverse of (1/x)+1 or 1/(x+1)? It's not exactly clear when typed haha!
But whatever the original equation is, remember when finding inverse you can equate any of the equations to x, because the inverse and the original graph ALWAYS intersects along the line y=x. Don't bother trying to equate them to each other.

As for implied domain of (x-1)/(x+2), IDK how others would do it but I would first convert it to -3/(x+2)+1, so from this you can tell that the asymptotes will be at y=1 and x=-2. Remember that the graph never touches the asymptotes, therefore you can deduce that the implied domain would be R\{-2}.

As for sketching, I usually start with the asymptotes, so draw two dotted lines for y=1 and x=-2. Then look at what shape the graph would be and whether or not it would have any x- or y-intercepts. Then draw the lines.

For graphing - it's good to know what shape each type of graph would be. Look at the power to know what type of graph it is. For example, in Q2 it's 16-x^2. A power of two implies it's a quadratic, which is a parabola. Then look at what form the graph is written in. For 16-x^2, you can rewrite this as (4-x)(4+x), so you can find out its x-intercepts. If it was written in ax^2 + bx + c, you can find the y-intercept. Finally, if it was a(x+b)^2 + c, you can find the turning point.

What I use to remember is that the power implies how many x-intercepts a graph could have up to. A linear (x^1) only has one intercept, quadratic (x^2) has up to two x-intercepts. A cubic (x^3) has up to three, and a quartic (x^4) would have up to four etc.

Hopefully that clears it up for you Lemme know if you have any more questions!
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#### Sine

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##### Re: Specific methods questions
« Reply #5 on: April 06, 2019, 11:38:48 pm »
+1
Also, did you just memorised the general shape and equation of every graph? I find it hard to draw a graph given an equation atm

I wouldn't say memorised - but over time when you continuously use the same graphs it will become entrenched in your memory. I would be going to the study design to see which graphs you actually need to know.

There are a few points I think are important when thinking about graphing
- A graph is just many points connected. Sure the number of points are infinite but if you just sub in values for x and see the output for y you should be able to graph after a certain number of points (this would be a last ditch effort to graph).
-A graph is just a simpler graph that has undertaken some "transformations" E.g. Say we have the graph y = 0.5sin(9-2x) + 2. A somewhat complex graph to sketch from scratch.
Consider it now in a different from.
y = 0.5sin(-2(x -9/2)) +2
(y - 2)/0.5 = sin(-2(x-9/2))
Now we may start with y = sin(x) which is a simple graph that you should know.
We can now just apply transformations until we get to the complex graph
y = sin(-x)
-
y=sin(-2x)
-
y/0.5 = sin(-2x) etc
By applying transformations one at a time it should become more manageable.

#### S_R_K

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##### Re: Specific methods questions
« Reply #6 on: April 07, 2019, 09:55:46 am »
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Hey! Sorry for my late response!

For Q1, can you please specify whether the question is asking for the inverse of (1/x)+1 or 1/(x+1)? It's not exactly clear when typed haha!
But whatever the original equation is, remember when finding inverse you can equate any of the equations to x, because the inverse and the original graph ALWAYS intersects along the line y=x. Don't bother trying to equate them to each other.

This is not true. Consider y = –x^3 and its inverse y = –x^(1/3). Their graphs intersect at points that are not on the line y = x.

The set of solutions to f(x) = f-1(x) will lie on the line y = x only if both functions are strictly increasing. Otherwise, that method does not work, and you will need to solve the equation f(x) = f–1(x).
« Last Edit: April 07, 2019, 01:22:41 pm by S_R_K »

#### DBA-144

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##### Re: Specific methods questions
« Reply #7 on: April 07, 2019, 10:39:01 am »
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This is not true. Consider y = –x^3 and its inverse y = –x^(1/3). Their graphs intersect at points that are not on the line y = x.

A function and its inverse will intersect along the line y = x if and only if both functions are strictly increasing. Otherwise, that method does not work, and you will need to solve the equation f(x) = f–1(x).

Would this mean that we need to state that "as f(x) is strictly increasing over (domain), then f(x)=f^-1(x) <--> f(x)=x"? Or are we allowed to go right to f(x)=x ?

#### S_R_K

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##### Re: Specific methods questions
« Reply #8 on: April 07, 2019, 12:30:26 pm »
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Would this mean that we need to state that "as f(x) is strictly increasing over (domain), then f(x)=f^-1(x) <--> f(x)=x"? Or are we allowed to go right to f(x)=x ?

As long as the assumption is true, I don't think you'd need to explicitly state it as part of your working (there's no evidence from examiners reports that it's required). I'd just do a rough sketch of the graph to confirm the assumption, and then jump straight into solving f(x) = f-1(x) using that proxy equation.

It's probably also worth noting that I've not seen an exam question where students are required to solve f(x) = f-1(x) where f is not strictly increasing. This is what leads to students forming the incorrect belief that this equation can be solved by solving f(x) = x. I can see this being something that examiners will target in the future.

#### AlphaZero

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##### Re: Specific methods questions
« Reply #9 on: April 07, 2019, 01:00:40 pm »
+2
Just want to clear a few things up.

First, it is NOT true that a function $f$ and its inverse $f^{-1}$ will always intersect on the line $y=x$.

They can for example not intersect at all:  $f(x)=e^x\implies f^{-1}(x)=\log_e(x)$.

They can also intersect at points that are not on $y=x$:   $f(x)=\sqrt{1-x}\implies f^{-1}:[0,\infty)\to\mathbb{R},\ f^{-1}(x)=1-x^2$.

Even worse, they can also intersect but never on $y=x$:   $f(x)=\dfrac{-1}{x}=f^{-1}(x)$.

In this thread we've also discussed what occurs when $f$ is strictly increasing. The correct statement relating points of intersection and strictly increasing is: $\boxed{f\ \text{is strictly increasing}\implies \text{any points of intersection of }f\ \text{with }f^{-1}\ \mathbf{must}\ \text{lie on }y=x}$ (Note that  $f\ \text{is strictly increasing}\implies f^{-1}\ \text{is strictly increasing}$)
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#### Justanotherhuman

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##### Re: Specific methods questions
« Reply #10 on: April 07, 2019, 07:41:27 pm »
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Hey guys!
If you could help me out with some more questions it'll be amazing!

- I'm having trouble separating dilations from y axis from x axis. I understand that it's about stretching and shrinking but have a look at the question below:
Determine the dilation when graph of y= root (5x) is obtained by dilating the graph of y= root x:
a) From y axis:
(x,y) -> (5x,y)
5x' -> x
x' -> x/5
That indicates a dilation of factor 1/5 from the y axis

b) from the y axis: the answer to this is root 5 and I have no idea why

- When stating how a graph has been transformed from 1 graph to another, how do you know what order to write it in (lol i know some people literally can visualise translation - not sure how to learn to do that..practice???). The DRT thing doesn't seem to always work.
a) State sequence of transformation that transforms graph of 1st equation to 2nd equation
y= 2/(3-x) + 4 , y= 1/x
How would you do that?

- While sketching the graph of y= 1-(x+4)^4
There are 2 x intercepts
1 = (x+4)^4
-3 = x
How do you find the other x intercept (which is at -5?)

- How is y = (3x+2)/(x+1) equivalent to 3 - (1))/(x+1)
How does it algebraically get there?

Thank you

#### colline

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##### Re: Specific methods questions
« Reply #11 on: April 07, 2019, 11:08:38 pm »
+1
@S_R_K @AlphaZero Didn't know about that at all! Thanks so much for pointing it out I don't want to accidentally spread around false info haha. Will definitely keep that in mind!

@Justanotherhuman I'm not 100% confident on your first two questions so I'll leave them for someone more mathematically capable As for the last two questions:

Quote
- While sketching the graph of y= 1-(x+4)^4
There are 2 x intercepts
1 = (x+4)^4
-3 = x
How do you find the other x intercept (which is at -5?)
Remember two negative numbers multiplied together makes a positive.
So the working out would look like this:

1 - (x+4)^4 = 0
(x+4)^4 = 1
(x+4) = +/- 1
x = 1 - 4 = -3 OR
x = -1 - 4 = -5

Quote
How is y = (3x+2)/(x+1) equivalent to 3 - (1))/(x+1)
How does it algebraically get there?
You can use long division. Tbh I find it very time consuming so this is my usual working out:

(3x+2)/(x+1)
= [3(x+1)-1]/(x+1)
= 3 - (1)/(x+1)

Thought process:
In order to simplify it, you must write your numerator in such a way that you can cancel out common factors.
So, instead of (3x+2), you should see how you can rewrite it so you get an (x+1) on the top.
(3x+2) can be rewritten as [3(x+1) - 1]
From this, you can then cancel out the common factor (x+1).

Hope that makes sense!
« Last Edit: April 08, 2019, 01:25:40 am by colline »
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