Problem Solving Strategies

Abcd

5:09 AM

@JohnRennie Are you there

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie morn, hi. Please see:

in Mathematics, 2 mins ago, by Abcd

in Mathematics, 2 mins ago, by Abcd

Can someone please explain point (ii)

in Mathematics, 2 mins ago, by Abcd

It says: if A is singular matrix and (adj A) D = O then the system of equations given by AX = D is consistent with infinitely many solutions. Why is that?

John Rennie

Linear algebra?

Abcd

Its for finding solution of:
$a_1 x + b_1y + c_1 x = d_1$, $a_2x....$ and till $a_3 x....$

@JohnRennie Just multiply LHS and you will get this system from definition of equality of matrices^

John Rennie

Aha, yes, the technique is often introduced as a way of finding solutions to simultaneous equations. Typically students aren't told that this is their first exposure to the subject of linear algebra.

Abcd

5:14 AM

@JohnRennie Oh okay. I dont know what linear algebra is.

John Rennie

@Abcd it doesn't matter anyhow ...

If you have a set of simultaneous equations then they may have one solution, no solution or infinitely many solutions.

Abcd

@JohnRennie Okay, can you explain point 2?

@JohnRennie yes

John Rennie

The solution to simultaneous equations is actually a matrix operation.

Abcd

ikr

John Rennie

The conditions ii and iii are conditions on the matrices involved in the solution that tell whether the equations have one, none, or infinitely many solutions.

Abcd

5:17 AM

ikr

John Rennie

Are you asking for a proof of point ii?

Abcd

@JohnRennie Yes, I want to know how he got that.

John Rennie

Offhand I don't know the proof. It's the sort of thing you learn for exams then forget because you never need it again.

Abcd

@JohnRennie i cant find it anywhere online too.

John Rennie

Do you need to know the proof? Obviously if $\det(A)=0$ everything blows up so there's no single solution.

Abcd

5:27 AM

@JohnRennie there some significance of adjoint here

John Rennie

@Abcd not really. It's just that $A^{-1} = adj(A)/|A|$

Abcd

@JohnRennie Does $\text{adj(A)D}= O $ make any physical sense to you?? (its given in point 2)

John Rennie

$\text{adj(A)D}= 0$ makes the right hand side $0/0$ and that is indeterminate, hence there are infinitely many solutions. Physically it means that two (or more) of your simultaneous equations aren't independent.

Abcd

@JohnRennie which RHS?

John Rennie

$ X = A^{-1}D$

The authour splits $A^{-1}$ into $A^{-1} = adj(A)/|A|$

$$ X = \frac{adj(A)D}{|A|} $$

Abcd

5:33 AM

Sorry wasnt paying attention to det(A)=0

@JohnRennie First of all inverse doesn't exist so its illegal to multiply.

$AX = D$ should be considered.

John Rennie

The adjoint always exists whether A is singular or not.

So you can always multiply $adj(A)$ and $D$

The only problem comes when you divide by the determinant.

The point is that if the determinant is zero you get problems, but you have to distinguish between $0/0$ and $x/0$ (for some non-zero $x$)

$0/0$ is indeterminate while $x/0$ is undefined.

Abcd

Wait please.

AX = D.

How does adoint come in the story?

John Rennie

The inverse is the adjoint divided by the determinant

Abcd

3 mins ago, by Abcd

@JohnRennie First of all inverse doesn't exist so its illegal to multiply.

@JohnRennie We are talking about singular matrix

inverse doesnt exist

John Rennie

OK we have a matter of definitions here I think

The inverse is always $adj(A)/|A|$

When we say the inverse doesn't exist that means $|A|=0$ and of course you cannot divide by zero

But there's nothing wrong with splitting the inverse into the adjoint and determinant and postponing the question of whether $|A|=0$ or not until later.

Abcd

5:43 AM

@JohnRennie Okay but in point (ii) we have $X = \text{null matrix}/0$ not $0/0$

John Rennie

In (ii) you get a matrix whose elements are all equal to $0/0$ i.e. a matrix all of whose elements are undetermined

Abcd

Yeah got it.

Got it, thanks!

Basically in 3 all solutions tend to infinity?

implies no solution.

John Rennie

Yes

Abcd

@JohnRennie is this faster then Cramer's rule?

John Rennie

@Abcd it is effectively what you're doing when you use Cramer's rule.

In practice for small equations (2 or 3 variables) I'd probably just solve them by hand, but in physics we often get massive sets of equations and have to solve them by computer. That algorithm will use the approach decribed in your book not Cramer's rule.

Nehal Samee

6:35 AM

@sammygerbil I thought so.

Nehal Samee

6:55 AM

@JohnRennie. There?

I wanted to know what amount of weight would act during extension of a rod hanging freely from a ceiling due to its own weight?

John Rennie

@NehalSamee morning :-)

Nehal Samee

@JohnRennie morning.. :)

John Rennie

The stretch isn't uniform i.e. the bits of the rod near the ceiling are stretched more than the bits of the rod near the lower end.

This makes sense because the bits of the rod near the ceiling have almost the whole mass of the rod below them and stretching them.

Nehal Samee

@JohnRennie Why is that so? I only know that tension is more at the upper end

John Rennie

While the bits of the rod near the end have almost no mass below them pulling them down.

Nehal Samee

6:58 AM

@JohnRennie OK

John Rennie

What is it you want to know? How to calculate the extension of the rod?

Nehal Samee

So.. What amount of original weight would act?

@JohnRennie I heard its 1/3 ..how?

John Rennie

@NehalSamee have a look at Sammy's answer here:

3

Q: Why does different part of a spring having mass expand proportional to their distance while the spring has some mass hanged in the bottom?

Why do different parts of a massive spring expand proportional to their distance while the spring has some mass hung on the bottom which is comparably very less than the mass of the spring? If we take a massless (or practically negligible mass) spring then the elongation of the spring is uniform....

Nehal Samee

@JohnRennie this case is dynamic extension as Sammy said, but what's static extension... Is it the horizontal one?

John Rennie

I'm not sure why Sammy uses the terms dynamic extension and static extension.

It's just the difference between the spring being stretched by its own weight, i.e. hanging vertically, and not being stretched by its own weight, i.e. resting on a horizontal surface.

Nehal Samee

7:13 AM

@JohnRennie but masses considered are different... M/3 and M/2 ...

John Rennie

Ah, OK, actually Sammy's answer is more general that I realised at a first glance.

If you have a massive spring hanging vertically there are two situations to consider.

If the spring of mass $m$ and force constant $k$ is just hanging stationary then its extension is given by $mg/(2k)$ i.e. the extension is half of what you'd get if you hung a mass $m$ on the end of a weightless spring.

But suppose you set the spring oscillating. In that case the mass of the spring has two effects. Firstly it creates the downward force on the spring due to gravity, and secondly it affects the acceleration of the various parts of the spring due to Newton's second law.

So the oscillating case is more complicated than the static case.

Abcd

7:47 AM

@JohnRennie Are you there

John Rennie

@Abcd hi

Abcd

@JohnRennie reflection matrix for reflection about y =x tan theta

how to find it

John Rennie

As I recall when you're trying to construct a transformation matrix you do it by considering what happens to the unit vectors (1,0) and (0,1)

The vector (1,0) gets rotated by an angle $2\theta$ so it would go to $(\cos 2\theta, \sin 2\theta)$

and the vector (1,0) gets rotated clockwise (negative angle) by $90-\theta$, so it goes to, erm, I need to draw this ...

Abcd

@JohnRennie Do you mean $\hat i$ and $\hat j$ respectively

John Rennie

Yes

Just draw the diagram and it's obvious ...

I can draw you a diagram to show what I mean if you want ...

Abcd

7:55 AM

yes

John Rennie

OK, give me a moment ...

Abcd

@JohnRennie I dont get this first.

John Rennie

The dashed line is the line $y = x\tan\theta$

So reflection in that line rotates (1,0) by an angle $2\theta$

Abcd

1 min ago, by Abcd

@JohnRennie I dont get this first.

@JohnRennie Please see this . I dont get why we have to consider 1,0 and 0,1

John Rennie

Suppose you have some transformation matrix $( (a,b), (c,d) )$

When you multiply it by $(1,0)$ the result is $(a,b)$

So the top row of the transformation matrix is the same as the point that $(1,0)$ maps to.

Abcd

8:05 AM

@JohnRennie no its a,c

If a b is first row. And c d is second row

John Rennie

Hmm, yes, maybe I mixed up my rows and columns.

But the basic principle remains true. By considering what the transformation does to (1,0) and (0,1) you can construct the transformation matrix.

Hmm, yes, it is the columns not the rows. Blame aging memory.

$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} a \\ c \end{matrix} \right) $$

$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} b \\ d \end{matrix} \right) $$

Abcd

@JohnRennie So what if we are getting a c and b d back?

John Rennie

If you look at my diagram, $(1,0)$ transforms to $(\cos2\theta, \sin2\theta)$

So we immediately know that the transformation matrix for reflection in $y=x\tan\theta$ has $a = \cos2\theta$ and $c=\sin2\theta$.

Abcd

3 mins ago, by Abcd

@JohnRennie So what if we are getting a c and b d back?

John Rennie

@Abcd I don't understand what you are asking

Abcd

8:15 AM

Suppose we have a point x and y.

And we reflect it in y = x tan theta

So you are saying consider the unit vectors of x and y?

Then transform them?

Then multiply To get new position?

John Rennie

No. All we are asked to do is find the transformation matrix. The matrix doesn't depend on the values of $x$ and $y$ (obviously) so if we can find the matrix for some specific choice of $x$ and $y$ we have found the matrix.

Abcd

just a min

Let me show from my textbook

John Rennie

What I'm saying is that if we consider the specific points (1,0) and (0,1) we can easily find the matrix

Abcd

John Rennie

That's what I'm trying to explain, but you don't seem to want to follow my explanation.

I have no idea what the authors procedure is. He doesn't give any explanation for how he derived his result.

Abcd

8:23 AM

Yes... that's what confused me there

John Rennie

What I'm explaining is a guaranteed way to get the trasnformation matrix.

Abcd

Okay, I will ignore this idiot author.

@JohnRennie please explain the use of this part^

John Rennie

@Abcd which part?

Abcd

@JohnRennie the message I have replied to. What is the use of doing that?

I mean we are getting a and c back in a new matrix, so how does that help?

John Rennie

We have some transformation matrix $((a,b),(c,d))$ where all of $a$, $b$, $c$ and $d$ are unknowns. Yes?

Abcd

8:27 AM

What is a transformation matrix really?

(no definition in my book)

John Rennie

Suppose you have some transformation, e.g. reflection or rotation, that takes the point $(x,y)$ to $(X,Y)$. Then we can write the transformation as the matrix equation:

$$ Ax = X $$

Where $x = (x,y)$ and $X=(X,Y)$ (sorry slightly rubbish choice of notation).

Or if we write this out in full:

$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} X \\ Y \end{matrix} \right) $$

The matrix $A$ is the matrix that describes the transformation, and that's what I am calling the transformation matrix.

Abcd

@JohnRennie I see, then?

@JohnRennie yes

John Rennie

And our task is to find the numbers $a$, $b$, $c$ and $d$.

Abcd

yes

John Rennie

Now suppose we work out where the transformtion takes the point $(1,0)$. We work this out by whatever construction is appropriate. This is usually easy because it's usually easy to work out where $(1,0)$ goes. OK so far?

Abcd

8:33 AM

yes

John Rennie

Suppose we find that $(1,0)$ maps to $(X,Y)$. Then we know that:

$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} X \\ Y \end{matrix} \right) $$

Abcd

ya

John Rennie

If we hand multiply the matrices we get $X=a$ and $Y=c$

Abcd

yes

John Rennie

So if we know $X$ and $Y$ that means we have found the first column of our transformation matrix.

Abcd

8:35 AM

Ya

John Rennie

35 mins ago, by John Rennie

Abcd

ya

X = cos 2theta

Y = sin 2 theta

So we get a and c.

Then?

John Rennie

Yes

OK now we do the same thing for the point $(0,1)$. We figure out where this point goes under the transformation. I can give you the answer or you can have a go at working it out ...

Abcd

jam

@JohnRennie I am getting sin 2 theta , -cos 2 theta

John Rennie

@Abcd yes, so now you know the second column of your transformation matrix.

i.e. you now know the values of $b$ and $d$

Abcd

8:46 AM

@JohnRennie Got it thanks!

@JohnRennie For how much time are you here

John Rennie

This is a general technique that works for all linear transformations

@Abcd I'm around for at least a couple of hours

Abcd

@JohnRennie ya its brilliant. Nothing like this is given in my book

John Rennie

@Abcd I'm a bit surprised because it is so simple to use

Abcd

@JohnRennie But there's an assumption. How can we assume that every transformation can be represented by those matrices?

John Rennie

Offhand I can't remember the proof, but I don't think it's hard. The key assumption is that the transformation is linear. I'd have to Google to find the details.

Linear map

In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (including vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication. An important special case is when V = W, in which case the map is called a linear operator, or an endomorphism of V. Sometimes the term linear function has the same meaning as linear map, while in analytic geometry it does not. A linear map always maps linear subspaces onto linear subspaces (possibly of a lowe...

See the Matrices section of that article

Ah I see Wikipedia has an extensive description of transformation matrices.

Abcd

8:53 AM

@JohnRennie thats not high school level stuff

@JohnRennie Can you tell me the simple reason?

John Rennie

@Abcd agreed, that's taking you down the rabbit hole of linear algebra. It's a fascinating subject but you don't need it for JEE you just need to know the main results.

@Abcd I don't think there is a simpler explanation than the one in the Linear map article. For now just accept that it's true and move on.

harambe

9:24 AM

@JohnRennie good morning

John Rennie

@harambe morning :-)

harambe

@JohnRennie imgur.com/a/wCATqIu

In water would there will be additional boyant force that we have to consider here?

But how will it help in the FBD... I mean it is not influencing the angle or the electric charge right?

John Rennie

Yes. The increased dielectric constant reduces the electrostatic force so the spheres would tend to hang more nearly straight down. However at the same time the buoyant force decreases the downwards force on the spheres and that compensates for the decrease in the electrostatic force.

harambe

Okay. Got it

John Rennie

Suppose we write the electrostatic force as $F_e$ and the net vertical force ($mg$ minus the buoyant force) as $F_v$, than the tan of the angle is $F_e/F_v$.

harambe

9:31 AM

Yeah

John Rennie

And you're told that this stays constant, so the fractional decrease in $F_e$ must be equal to the fractional decrease in $F_v$.

harambe

Okay

@JohnRennie About the angle.... I think I am confused. I thought the angle between the horizobtal and vertical force is 90..the electric force horizobtally and the Fv vertical ly

Never mind. I got it

John Rennie

Cool :-)

harambe

9:47 AM

@JohnRennie got the answer. Your method looked way clean than the solution showed

John Rennie

@harambe naturally :-)

harambe

XD

@JohnRennie if two spheres having say charges Q1 and Q2 are touched then the resultant charge on them will be Q1+Q2/2 on each

Would this mean electric force will change too

John Rennie

Well the force is initially $kQ_1Q_2/r^2$ and after they have touched it is $k(Q_1+Q_2)^2/(4r^2)$

If you set the two forces equal you'll find it is only true for specific values of $Q_1$ and $Q_2$

harambe

That's why... I had this question in a test and I just wrote it will be greater than the initial force but it was said it could be equal too....

harambe

10:12 AM

@JohnRennie imgur.com/a/KbB73D0

My expression for restoring force is: -kq^2y/2(a^2+y^2)^3/2

How do I proceed from here

I have ignored mg because its a constant force here

John Rennie

Use a binomial expansion.

harambe

I think there is some binomial series after this but how to use it

Should I only include the first two terms of the BT?

Will others be negligible

John Rennie

The $y \ll a$ is a dead giveaway. Expand the force as a power series in $y/a$ and discard terms in $(y/a)^2$ and higher.

harambe

Got it

John Rennie

Is there a mistake in that question? Should it be two charges at $x =+a$ and $x=-a$. That would make more sense

harambe

10:22 AM

@JohnRennie yes. I check it with the original question of jee and this mistake is rectified there so thee is a misprint here

John Rennie

Thought so :-)

Ah, no.

harambe

I am getting $y^3$

John Rennie

It's (a)

harambe

It doesn't match the options.....

@JohnRennie yes

John Rennie

It can't be c or d because the force is zero when $y=0$

And if you move in the positive $y$ direction the force is positive, so it has to be $+y$ not $-y$.

I can have a go at doing the calculation in bit, but right now I'm tied up in an online chat thing.

harambe

10:28 AM

Okay

I can do other questions meanwhile

John Rennie

11:07 AM

@harambe actually I don't think you need a binomial expansion. It's simpler than that.

harambe

Okay. Can you tell me about it

John Rennie

Let me draw a diagram ...

$$ F =\frac{k Q/2 Q}{r^2} $$

The vertical component is $F\sin\theta = F (y/r)$. And double that because we have a force from both sides. So we end up with:

$$ F_v =\frac{k Q^2 y}{r^3} $$

@harambe OK so far?

harambe

Yeah

John Rennie

And $r^2 = a^2 + y^2$ so we get:

$$ F_v =\frac{k Q^2 y}{(a^2 + y^2)^{3/2}} $$

harambe

Okay

John Rennie

11:19 AM

And if $y \ll a$ then $a^2 + y^2 \approx a^2$ so we just get:

$$ F_v \approx \frac{k Q^2 y}{a^3} $$

Or $F_v \propto y$

harambe

Got it. So we neglect y/ a fraction

@JohnRennie when we keep two spheres in contact then I read that charges will redistribute.... Why does this happen

John Rennie

It assumes the spheres are conductors. They don't have to be good conductors, just as long as they aren't perfect insulators.

When you touch two conductors together you effectively have a single conductor. And charge always distributes itself over the surface of a conductor.

If the balls are the same size then the symmetry means you'll end up with the same charge on each ball.

harambe

Okay

Can you help me with a similiar question based on this

imgur.com/a/5vXP1Hc

John Rennie

Question 9?

harambe

11:35 AM

Yeah. About the spheres one l

John Rennie

OK, when you connect the two spheres their voltage has to be the same. Yes?

harambe

eh Voltage .... How is it a factor here

John Rennie

@harambe Because the charge on an object is related to the voltage and capacitance of the object by the usual $Q = CV$.

Hmm, actually we could do the problem without needing to invoke capacitance. Maybe that would be more intuitive.

harambe

Okay. It will be better if I could understand the method intuitively so that I can remember it easily

John Rennie

Suppose the charge on the first sphere is $Q_1$. Then the electrostatic potential at the surface of the sphere is $V = kQ_1/R$. OK so far?

harambe

11:42 AM

Ok

John Rennie

And likewise for the second sphere we get $V = kQ_2/(3R)$

harambe

Okay

John Rennie

The thing is that we've connected the two spheres with a conducting wire, so electrons can freely move along the wire from one sphere to the other. At equilibrium the potential has to be the same at both ends of the wire otherwise charge would keep flowing along the wire until the potentials became the same.

And that means $V_1 = V_2$

harambe

Okay

John Rennie

So that means:

$$ \frac{kQ_1}{R} = \frac{kQ_2}{3R} $$

$$ Q_2 = 3Q_1 $$

harambe

11:47 AM

Okay and now we can calculate the electric field of the spheres

One doubt

John Rennie

@harambe Yes?

harambe

When you said potential then about which point were you talking potential about

On the sphere

John Rennie

If you're outside a spherically symmetric charge then it behaves like a point charge at the centre of the sphere. Yes?

harambe

Yea.

For points outside the sphere

John Rennie

So at a distance $r \ge R$ from the first sphere the potential energy is just the usual $kQ/r$.

harambe

11:51 AM

Oh so you were talking about the portion of wire attached to the spherical surface?

John Rennie

The wire is connected to the surfaces of the spheres

harambe

Got it

Nehal Samee

12:05 PM

@JohnRennie that means... If the spring is static in vertical position without mass, then half of weight acts for extension... If it's oscillating with/without mass, then one third of weight acts... Right?

harambe

1:51 PM

@JohnRennie are you interested in about maximum / minimum of functions

John Rennie

@harambe do you mean can I asnwer questions on it? If so, yes, I'm willing to give it a go.

harambe

Awesome. Yes

@JohnRennie imgur.com/a/478XVNc

I need to find the range if this function

So I have been taught that I have to differentiate this function, find the critical points. That will give me local macima/ minima

John Rennie

OK, so if there is no maximum or minimum in the domain $x=0$ to $3$ then then the range is just from $f(0)$ to $f(3)$. Yes?

harambe

why?

John Rennie

There are only three cases:

1. the function changes smoothly from $x=0 \to 3$. In that case the maximum and minimum values are at the ends of the domain i.e. at $x=0$ and $x-=3$. Yes

harambe

2:02 PM

What do you mean by " changes smoothly "

John Rennie

Let me draw a diagram ...

I've drawn two functions. The red curve increases monotonically in the range 0-3 and green decreases monotonically in the range.

In both cases the maximium and minimum values are at the ends of the range 0-3.

harambe

Okay

John Rennie

But case 2 is when the function has a maximum:

In that case the minimum value is at one of the ends of 0-3, but the maximum is not in between at some value we have to find.

And I won't draw it, but case 3 is if the function has a minimum in 0-3

So we need to find out if our function has any maxima or minima in the range $x = 0 \to 3$

harambe

So In case 2 we know the minimum is at one of the end points but not the maximum... It can be anywhere between them

John Rennie

Correct. So we need to evaluate $f(x)$ at both the end points and at the maximum

harambe

2:12 PM

Okay

John Rennie

Do you know how to find the maxima and minima of a function?

harambe

Yea

John Rennie

So go ahead. Find the values of $x$ for the maxima and minima.

harambe

At maxima and minima the values are -14 and 13

At end values of function at 0 and 3 the values are 6 and -3

So range comes out to be (-14, 3)

Does this look right

John Rennie

Aren't you only considering the interval 0 to 3?

harambe

2:21 PM

Oops... (-14, 13)

John Rennie

That's your function in the interval 0 to 3

harambe

Okay.... Let me try again. Btw what is the website for this graph plotter

John Rennie

I used Microsoft Excel

harambe

The range looks like [-14, 6]

John Rennie

Yes

harambe

2:26 PM

I need to get that too.

I got -14

John Rennie

You can use Google Sheets. For simple stuff that is as good as Excel.

@harambe The function has a maximum at $x=-1$ and a minimum at $x=2$. Yes?

harambe

Yea

I am getting difficult y in maximum part

John Rennie

So we need to evaluate $f(0)$, $f(3)$ and $f(2)$ to be sure of covering the whole range.

harambe

Yea

John Rennie

We don't care about $x=-1$ because it's outside our interval of 0-3

Fawad

2:30 PM

@harambe are preparing for jee?

harambe

Oh yea.. That why I was getting the wrong answer

@Fawad yes unfortunately

Fawad

You are from cbse?

harambe

Yes

@JohnRennie is the steps combination of case 1 and case 2 or just case 2

I mean the steps of solution here

John Rennie

This is case 3 i.e. it has a minimum within the interval we are looking at.

harambe

So we do the solution and look for the case the solution falls into?

John Rennie

2:35 PM

First find the maxima and minima.

Fawad

@JohnRennie how do you prove that image of orthocentre for a triangle lies on cirmumcentre of that triangle with respect to any side?

John Rennie

Then if the $x$ position of the maxima and minima lie outside the interval there is nothing more to do. Just evaluate $f(x) at the end points of the interval.

@Fawad no idea, I would have to Google it.

@harambe does that make sense so far?

harambe

Yea

Fawad

2

Q: Why would the reflections of the orthocentre lie on the circumcircle?

Let ABC be a triangle which it not right-angled. Define a sequence of triangles $A_iB_iC_i$,with $i \geq 0$, as follows: $A_0B_0C_0$ is the triangle $ABC$; and, for $i \geq 0$, $A_{i+1}$, $B_{i+1}$, and $C_{i+1}$ are the reflections of the orthocentre of triangle $A_iB_iC_i$ in the sides $B_iC_...

geometry contest-math

John Rennie

@harambe If there are any maxima or minima in the interval then we need to find them and evaluate $f(x)$ at all the maxima and minima as well.

harambe

2:38 PM

Okay

John Rennie

In this case there was just a minimum in the interval.

harambe

Yea

John Rennie

But if the interval had been -2 to +3 then there would have been both a minimum and a maximum so we'd need $f(x)$ at both those values as well as at the ends of the interval.

harambe

Got it but if the question had one of the end points in open bracket ie they were not in domain then how can we proceed there?

John Rennie

@harambe not sure to be honest. Suppose in this case 0 and 3 weren't in the domain. In that case I guess you'd have write the upper limit of the range a $6 - \epsilon$, where $\epsilon$ is arbitrarily small.

To be honest my grasp of the more formal end of maths isn't good enough for me to know what the answer is.

harambe

2:48 PM

Oh kay. It's no problem. I will ask this on the math chat

Fawad

@JohnRennie if domain is [0,3) then range is written as [-14,6) (if f(3)=6)

John Rennie

@Fawad Yes, that makes sense. @harambe

Fawad

@JohnRennie but max is not given by extreme hmmm

harambe

I think I get it. If the function is continious the finding the limit at both points will give the value of function there maybe?

Yea. Continious function can flactuate but momotonous function can't. But still finding limit wont hurt right

Fawad

I knew it was dumb comment so I deleted it

John Rennie

2:55 PM

@harambe You always need the values of $f(x)$ at the ends of the interval. Just because there is a maximum and/or minimum in the interval doesn't mean those are the highest/lowest values of the function in the interval.

Suppose in your question you had been given the interval -100 to +100. In that case the range is from $f(-100)$ to $f(100)$ because those are lower and higher than the values at the minimum and maximum.

harambe

Why does this happen only at end points

Why do we check only for end points

Because they are influencing how the graphs can be made?

Cases to be exact

John Rennie

If there is no maximum or minimum the function changes monotonically between the end point. Yes?

harambe

Yes

John Rennie

If there is a single maximum the function changes monotonically from the start point to the maximum, then monotonically from the maximum to the end point. Yes?

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