Math help (2 Viewers)

[Fawun]

Sy answered it though. What do you have troubles with?

Everything. I still don't get it.

If you haven't noticed yet, i'm a very slow learner. I have to read things a gazillion times to be able to comprehend and register it into my puny brain.

It will help later on.

In what?

 

[SpiralFlex]

Everything. I still don't get it.

If you haven't noticed yet, i'm a very slow learner. I have to read things a gazillion times to be able to comprehend and register it into my puny brain.


In what?

Which parts don't you get?

 

[HSC2014]

Okay so the question is basically saying this. I'll try to keep this as simple as possible.

There are 3 regions, each with a different function/graph style to it.
The critical points are -2, and 3. (You can determine this by looking at the domain conditions on the right. At these points the graphs change). So you can conclude that the 3 regions are.. on the left of -2 (x < -2), between -2 and 3 (-2 <= x <= 3), and on the right of 3 (x > 3).

So in each of these regions, there will be a different style graph. (I'm just wording out what the question is saying, to make sure you understand it).
When x is less than -2, y = 1 - x^2.
Between or equal to -2 and 3, y = x,
When x is greater than 3, y = 2x.

You can draw these and see where the sharp points/discontinues lie, and thus answering the question. Do it now!
With experience you can form these graphs in your head and it will be very easy to see the answers (Note that I haven't learnt differentiation and stuff like other people have mentioned in the thread, and this is just my personal way of doing it).

Edit: The question would be read like this (just in case you're lost in all the symbols and whatnot):
f(x) = 2x for/when x > 3,
3 when -2 <= x <= 3,
and 1 - x^2 when x > 3

Edit2: Oh the question is asking for any points that are not differentiable. Well then you'd just look at the boundaries of the regions (-2 and 3) where they transition smoothly and there is not a sharp change (Since at these points that is where change in graphs occur and can result in sharp/discontinuous points).

 

[Skycineia]

Spiralflex - I seek ur blood

Or

You could graph each one in pencil for all real x and truncate the graph at the appropriate regions. Then trace over it with pen

 

[SpiralFlex]

Spiralflex - I seek ur blood

Or

You could graph each one in pencil for all real x and truncate the graph at the appropriate regions. Then trace over it with pen

hi Aysce

 

[Sy123]

How do you know that there is a 'transition' between two functions through x=2 and x=-3?

Look at our piece-meal function and its components, we know that at x=2 and x=-3 there is a transition because they are the endpoints of the domains for the functions that connect to the other ones. That is, look at the first function for instance its domain is x > 3 so what this means is that at x=3 there is a transition between function to another because its an endpoint of a domain.
The middle function will give us both critical points since its the middle so its endpoints are both transition points.

Now I MUST stress the fact that not ever function has to cover all points of x, that is each critical point doesnt need to be a transition point (just in this case they are), these crticial poitns will be at the ends of every domain and thats what we need to test whether they are non differentiable or not

Do you always have to observe the gradient of these functions? or is it just this question that you have to do it?

Note here that Im only testing the gradient of the function because I want to be able to tell whether they connect smoothly or not.
The transition between two functions is smooth is only the case IF
-> They have the same gradient at the intersecting or transition point
-> They have the SAME y-value at that point.

If it satisfies both condition the transition is smooth and hence is differentiable at that point. I observed the gradient of the two lines, found that they are different gradients and hence I can assume that the transition between these two lines however it may occur will NOT be smooth hence wont be differentiable.

And what functions? There are like 3 functions in the question :s

A piece-wise function which is what you were given is one whole function composed of many different relations that are functions.
Anything can be a function if it satisfies having having only 1 y-value for every x-value It satisfies the vertical line test
A piecewise function is a function that is composed of different relations that happen to be functions.

So in a piece-wise function since it is a function composed of other functions, then each function must have a unique domain for which the function is true.
I'll explain this further when I respond to you asking where y=1-x^2 came from

How do you know that it connects the graph f(x)=2x and f(x)=3? Where did you even get x=3 anyway?

Poor choice of wording on my part, we dont know whether they connect at all, they may or they may not. The point is we are looking for whether they have the same gradients so that if they DO connect then if they have different gradients they MUST not be smooth, draw this out if its unclear. I got x=3 from the endpoints of the domain for the functions

Why are you only looking at y=1-x^2 now? why aren't you looking at the first two functions anymore?

Since a piece-wise function is a fucntion composed of many different relations, y=1-x^2 is one such relation and in this piece-wise function the relation is valid for the x in the domain. I am looking at this one because x=-2 is where we connect y=1-x^2 to y=3 (and seeing whether they connect smoothly)

Why are you questioning x=-2? Why -2?

Because since the parabola has a varying gradient we dont know FOR SURE whether they will have the same gradients at that point, so we must test whether it is. Since I assume you havent learnt what differentiation is Ive taken an opposite argument here.
After you learn calculus this is the argument that you will pose:

Is the gradient of the parabola at x=-2 the same as the gradient of y=3?

For now we ask ourselves:

Since y=3 has zero gradient does the graph connect with the part of the parabola that has a zero gradient, which is the turning point, the flat point

So we question x=-2 because we dont know whether they have the same gradient or not when they connect (once you learn calculus you will easily differentiate the parabola etc etc)

Wait you just said before that the turning point is x=0 which isn't x=-2. So whenever the turning point isn't the same as the point that we are questioning, it means that it has a sharp point?

ONLY because y=3 has a zero gradient so because our zero gradient of the parabola IS NOT x=-2 (it is x=0 the turning point), then we can say that they have different gradients at x=-2, whatever they might be we dont really need to care because the point is definitely not differentiable (whether sharp or discontinous or otherwise)



Now I should of said this before.

See the conditions of transition points I said before?
The transition between two functions is smooth is only the case IF
-> They have the same gradient at the intersecting or transition point
-> They have the SAME y-value at that point.

Now what we did before is test the gradients of the functions at this transition point. What you must note is that just because they have the same gradients, does not mean the curve is smooth. They may have different y-values at that point which will make the transition discontinuous.

Now we dont have to test gradients, we CAN test whether they have the same y-values, arguably testng y-values is easier in most cases but it requires substitution, for straight lines we can tell whether they have different gradients straight away and hence straight away tell whether any intersection if it occurs will be sharp or not.

 

[HSC2014]

If three non-zero real numbers, a, b, and c are found such that:
root(a+b) + root(b+c) = root(c +a)

a. Explain why b must be negative

I can't seem to figure it out. I had the answer a while back when I did it a few months ago but I guess that just means I didn't fully grasp it.

 

[Sy123]

If three non-zero real numbers, a, b, and c are found such that:
root(a+b) + root(b+c) = root(c +a)

a. Explain why b must be negative

I can't seem to figure it out. I had the answer a while back when I did it a few months ago but I guess that just means I didn't fully grasp it.

So we have the conditions:





Square both sides of our equation



Collect like terms, and divide everything by 2



Now this is important, we know that



That means if that square root is greater than zero, that means the b must be less than zero in order to 'balance' out the equation and make it equal to zero.
However take the trivial case of when the square root is equal to zero itself, IF that is true then b must be zero. However we are given specifically that they are all non-zero real numbers hence it applies for this case.

 

[HSC2014]

Oh wow thank you! I love that question 8) What can't you do haha

 

[Fawun]

these crticial poitns will be at the ends of every domain and thats what we need to test whether they are non differentiable or not

So the critical points are the numbers at the ends of every domain? So in this case, it's -2 and 3?

Note here that Im only testing the gradient of the function because I want to be able to tell whether they connect smoothly or not.

I don't get it. You're confusing me :s how do you test the gradient of the function?

The transition between two functions is smooth is only the case IF
-> They have the same gradient at the intersecting or transition point
-> They have the SAME y-value at that point.

And how do you know if they have the same y-value? How do you test it?

If it satisfies both condition the transition is smooth and hence is differentiable at that point. I observed the gradient of the two lines, found that they are different gradients and hence I can assume that the transition between these two lines however it may occur will NOT be smooth hence wont be differentiable.

So what you're saying is that it won't be differentiable because it has different gradients?

Because since the parabola has a varying gradient we dont know FOR SURE whether they will have the same gradients at that point, so we must test whether it is.

How do you test it?

So we question x=-2 because we dont know whether they have the same gradient or not when they connect (once you learn calculus you will easily differentiate the parabola etc etc)

Just to clarify before I ask "why aren't you testing x=3 anymore", you're testing x=-2 because it's the domain of the parabola right?

ONLY because y=3 has a zero gradient so because our zero gradient of the parabola IS NOT x=-2 (it is x=0 the turning point), then we can say that they have different gradients at x=-2, whatever they might be we dont really need to care because the point is definitely not differentiable (whether sharp or discontinous or otherwise)

So if they have different gradients, it's discontinuous?

Now what we did before is test the gradients of the functions at this transition point. What you must note is that just because they have the same gradients, does not mean the curve is smooth. They may have different y-values at that point which will make the transition discontinuous.

Okay once again just to clarify, you tested the gradients by putting them in the y=mx+b form right? and how do you know that they have different y-values?

Now we dont have to test gradients, we CAN test whether they have the same y-values, arguably testng y-values is easier in most cases but it requires substitution, for straight lines we can tell whether they have different gradients straight away and hence straight away tell whether any intersection if it occurs will be sharp or not.

Okay so if I were to be given another question similar to this, can I still find the answer without testing the gradients? and once again, how do you test the y-values?

Thanks again for taking the time to help me

 

[Fawun]

bump

 

[Sy123]

So the critical points are the numbers at the ends of every domain? So in this case, it's -2 and 3?



I don't get it. You're confusing me :s how do you test the gradient of the function?



And how do you know if they have the same y-value? How do you test it?



So what you're saying is that it won't be differentiable because it has different gradients?



How do you test it?



Just to clarify before I ask "why aren't you testing x=3 anymore", you're testing x=-2 because it's the domain of the parabola right?



So if they have different gradients, it's discontinuous?



Okay once again just to clarify, you tested the gradients by putting them in the y=mx+b form right? and how do you know that they have different y-values?



Okay so if I were to be given another question similar to this, can I still find the answer without testing the gradients? and once again, how do you test the y-values?

Thanks again for taking the time to help me

The approach to the question graphically

The logic behind gradients, smoothness and discontinuity, Note the the difference between the third and second case

What I mean when I say there is a possibility that the straight line intersecting with the curve turns out to be smooth, note turning point at x=1, what I was doing for x=-2 in YOUR question was testing whether this WAS an outcome (however we found that this is not the outcome)

You must note that it doesnt HAVE to connect (like the question you gave)

We test gradients of straight lines via y=mx+b, m is the graident
We test gradients of curves later using calculus (explored in depth once you learn calculus properly)
We test y-values via subbing in our numbers into the equation, for instance to find the y-value of 1-x^2 at x=-2 we sub it in and we find y=-3 as an output (as demonstrated on the graph I drew)

If I missed anything post what I missed

 

[Demento1]

Maths Help Thread - Home of Sy123 the tutor and his student, fawun.

 

[Fawun]

The approach to the question graphically

The logic behind gradients, smoothness and discontinuity, Note the the difference between the third and second case

What I mean when I say there is a possibility that the straight line intersecting with the curve turns out to be smooth, note turning point at x=1, what I was doing for x=-2 in YOUR question was testing whether this WAS an outcome (however we found that this is not the outcome)

You must note that it doesnt HAVE to connect (like the question you gave)

We test gradients of straight lines via y=mx+b, m is the graident
We test gradients of curves later using calculus (explored in depth once you learn calculus properly)
We test y-values via subbing in our numbers into the equation, for instance to find the y-value of 1-x^2 at x=-2 we sub it in and we find y=-3 as an output (as demonstrated on the graph I drew)

If I missed anything post what I missed

Okay thank you so much! It was a clear and good explanation!

Would you mind checking this question for me to see that i'm on the right track please? (can't be bothered latexing the whole thing up)

I don't know where to go next. I was thinking of factorising it but they have different powers :s

Would you be able to give me a hint as to where to go next?

 

[asianese]

WTF fawn. Sub x=1 immediately after differentiation.

 

[Fawun]

WTF fawn. Sub x=1 immediately after differentiation.

Yeah but I get some random number.

The answer is root 7 that's why I asked.

Derp, i know how to differentiate.

 

[asianese]

Isn't it ???

 

[deswa1]

Isn't it ???

This

 

[asianese]

And sorry Fawn i'm such a derp

 

[Fawun]

Isn't it ???

??

I don't know that's what the textbook says.

And sorry Fawn i'm such a derp

it's okay bby