Understanding the Use of the Long Jump Formula R=9.21m and Its Derivation

[rudransh verma]

Homework Statement

In 1991 Mike Powell jumped 8.95 m. Powells speed on takeoff was 9.5 m/s and g=9.8 m/s^2. How much less was powells range from the max possible range for a particle launched at same speed?

Relevant Equations

$$R={v_0}^2\sin(2\theta)/g$$

$$R=9.21 m$$
$$Difference = 9.21-8.95$$
$$D=0.26m$$

My question is when do we use the formula for R given above. Because we could have calculated the Rmax by $R=(v0\cos(\theta))t$ and then subtracted R from it to get the answer.
Why the x and y component in the derivation of this (R)formula are combined by eliminating t?

 

[anuttarasammyak]

R is x when y = 0 again.

 

[haruspex]

we could have calculated the Rmax by $R=(v0\cos(\theta))t$

How do you know what t would have been for max range?

Btw, the given formula is only for when the trajectory starts and finishes at the same height.

 

[SpectraPhy09]

Why the x and y component in the derivation of this (R)formula are combined by eliminating t?

Because we wanted to find the trajectory of the particle Which will be in form of its Y-Coordinate & X-Coordinate
You can use this formula also because for this situation when the Body is landing on the same height from where it started We can simply write Y=0 and then Find X(eliminating x=0 because that would be the starting point ).

 

[jbriggs444]

For a real world long jump, the landing position has the jumper's center of mass lower than at launch due to posture. One launches standing up. One lands as low as one can manage.

 

[kuruman]

For a real world long jump, the landing position has the jumper's center of mass lower than at launch due to posture. One launches standing up. One lands as low as one can manage.

Also, the length of the jump is from the takeoff line to the point where the jumper's first body part touches the sand. The feet are considerably far ahead of the center of mass and that is why jumpers try to land with feet as forward as possible without their butt hitting the sand first.

 

[jbriggs444]

Also, the length of the jump is from the takeoff line to the point where the jumper's first body part touches the sand. The feet are considerably far ahead of the center of mass and that is why jumpers try to land with feet as forward as possible without their butt hitting the sand first.

Or after.

"Long jumps are measured from the forward edge of the take-off board to the impression in the landing pit closest to the take-off board made by any part of the body of the jumper."

 

[kuruman]

Or after.

"Long jumps are measured from the forward edge of the take-off board to the impression in the landing pit closest to the take-off board made by any part of the body of the jumper."

That contradicts what I found here. Goes to show how reliable the web is.

 

[jbriggs444]

That contradicts what I found here. Goes to show how reliable the web is.

View attachment 292187

Adopting a "first point hit" marking policy would favor a strategy of laying out flat and, perhaps, bending the knees slightly to ensure a heel-first hit. This would also make measurement more prone to irresolvable dispute.

The reference I had found was here.

 

[rudransh verma]

How do you know what t would have been for max range?

We need to find the max R which can be calculated by putting $theta=45$ in this formula. I guess we have to use this formula anyway.
In the derivation of the formula they took two eqns for s, one vertical and horizontal and eliminated t. So the whole point of this is to find R easily even when t is not given.
$$R=v0cos(theta0)t$$
$$0=v0sin(theta0)t- \frac12gt^2$$
$$R=\frac{v0^2}{g}sin2(theta0)$$
We don’t know t and theta both, Yes?

 

[jbriggs444]

We don’t know t and theta both, Yes?

You already said that you know what theta needs to be to maximize R.

You had a perfectly good formula involving $v_0$ (known), $g$ (known), $\theta$ (known) and $t$ which you could easily solve for t:$$0=(v_0 \sin \theta)t - \frac12gt^2$$In fact, you did solve for $t$ and used the solution in your next equation.

[Minor edits on your $\TeX$. I used \sin instead of sin. This changes the font, removing italics, improving readability. I used \theta rather than theta0. The zero subscript was distracting and the Greek letter is shorter than the name of the Greek letter. I used the subscript introducer ("_") for the subscript on $v_0$. I thought about removing that subscript entirely, but left it to preserve similarity with the original]

 

[rudransh verma]

How do you know what t would have been for max range?

@jbriggs444 is right. I can find t and then I can use horizontal displacement eqn to get max R.

But the original doubt of this thread is final R eqn is used when you are not given t and launch angle $\theta$ is given.

 

[haruspex]

@jbriggs444 is right. I can find t and then I can use horizontal displacement eqn to get max R.

But the original doubt of this thread is final R eqn is used when you are not given t and launch angle $\theta$ is given.

In post #1 you are given v. You ask why you would use $R=v^2\sin(2\theta)/g$ to find $R_{max}$ "instead of using t".
The formula tells you that R is maximised at $\theta=\pi/4$, and using it you don't need to find t.
If you want to do it using t then you first have to find t. The formula you are using to do that requires you to know $\theta=\pi/4$. How did you know that except by using $R=v^2\sin(2\theta)/g$?

So I am baffled as to a) why you think it is simpler going via t instead of using the formula and b) why you think you have succeeded in doing so.

 

[haruspex]

But the original doubt of this thread is final R eqn is used when you are not given t and launch angle θ is given.

What are you doubtful about? Are you questioning whether the range formula is valid for all theta?

 

[rudransh verma]

You ask why you would use R=v2sin⁡(2θ)/g to find Rmax "instead of using t".

I am questioning the existence of this formula. When you can do it via the obvious way by finding t why have we derived a new formula?

The formula tells you that R is maximised at θ=π/4, and using it you don't need to find t to calculate R, just $\theta$

Ok so this is your answer. So it was not known experimentally first but by the formula first.

 

[haruspex]

I am questioning the existence of this formula. When you can do it via the obvious way by finding t why have we derived a new formula?

As I posted, we first have to find t without using $\theta=\pi/4$. We can do that by writing the two usual equations:
$y=v\sin(\theta)t-\frac 12gt^2$, $x=v\cos(\theta)t$
Substituting y=0 to find the time when it lands:
$2v\sin(\theta)t=gt^2$, $2v\sin(\theta)=gt$
Plugging that into the x equation:
$x=v\cos(\theta)2v\sin(\theta)/g=v^2\sin(2\theta)/g$
Now we can see x is maximised by $\theta=\pi/4$.

The formula exists. It is your choice whether to remember it or to remember how to deduce it. I was never good at remembering formulae, but I could usually remember how to get them.

A more challenging version is to consider max range on a slope - a standard calculation for gunners in days gone by.

 

[rudransh verma]

@haruspex I have the formula for $R=\frac {{v_0}^2 sin2 \theta}{g}$ and seeing it we conclude at $\theta=45$ we have max R.
But R is also equal to $R={v_0}cos\theta t$ and here R is max when $\theta=0$.
So what is it that I am missing?

 

[jbriggs444]

@haruspex I have the formula for $R=\frac {v0^2 sin2 /theta}{g}$ and seeing it we conclude at $\theta=45$ we have max R.
But R is also equal to $R=v0cos\theta t$ and here R is max when $\theta=0$.
So what is it that I am missing?

Let us begin by working with those $\LaTeX$ skills.

You wrote $R=\frac {v0^2 sin2 /theta}{g}$ which yields: $R=\frac {v0^2 sin2 /theta}{g}$

You wanted ##R=\frac{{v_0}^2 \sin 2 \theta}{g}$which yields:$R=\frac{{v_0}^2 \sin 2 \theta}{g}$Edit: Learned something myself. Putting squiggle brackets around {v_0} improves the look of the superscript for exponentiation -- moves it above and to the right of the whole$v_0$instead of putting it above and to the right of the$v$which ends up directly above the subscript$0$where it looks strange. So what is it that you are missing? You are missing that$t$(the time until vertical position returns to zero) is a function of$v_0$and$\theta$. You cannot set$\theta$to zero and expect$t$to remain unchanged. To put it in another way, you are working to maximize the function:$R(\theta) = v_0 \cos \theta\ t(\theta)$but you are erroneously treating$t$as a constant rather than as a function of$\theta##.

 

[kuruman]

What you are missing is that, on one hand if you set the two equal you get$$R=v_0 \cos\theta t=\frac{v_0^2\sin(2\theta)}{g}=2\frac{v_0^2\sin\theta\cos\theta}{g}\implies t=\frac{2v_0\sin\theta}{g}.$$On the other hand, if you solve the vertical motion equation for the time of flight, you get $$0=v_0 t\sin\theta -\frac{1}{2}gt^2=t\left(v_0 \sin\theta -\frac{1}{2}gt\right)\implies t=\frac{2v_0\sin\theta}{g}.$$There is no contradiction between the two equations.

 

[Steve4Physics]

So what is it that I am missing?

I suspect what you are missing is this...

There are often several correct methods to solve a problem. Which method you choose depends on identifying the different methods and spotting which method requires the least effort. The ability to do this depends on understanding and experience.

If done correctly, each alternative method will give the same answer.

Some calculations are more common than others so we use pre-derived formulae for convenience. For example, we often use the formula $s=ut + \frac 1 2 a t^2$. But we don't have to – we could solve the problem starting with the definitions of speed and acceleration; this would take more work.

The same goes for the range formula. Calculations involving (horizontal) range are common. We could do the calculations without using the range formula if required., But using the range formula (if permitted) might save us some work.

 

[rudransh verma]

To put it in another way, you are working to maximize the function: R(θ)=v0cos⁡θ t(θ) but you are erroneously treating t as a constant rather than as a function of θ.

R is max at 45. That we have deduced. How can we show it here using this formula of R.
All I know is $cos\theta$ is max at zero.

 

[jbriggs444]

R is max at 45. That we have deduced. How can we show it here using this formula of R.

As I understand it, you want to erase from our minds the already deduced information that R is maximized when $\theta$ is 45 degrees. Then you want to demonstrate that R is maximized at 45 degrees based on the formula: $R(\theta)=v_0 \cos \theta\ t$.

The straightforward approach to rewrite $t$ as a function of $\theta$ [which, as @kuruman points out has already been done upthread] and then solve for the $\theta$ that maximizes the resulting formula.

This is the standard crank-and-grind approach that you should try when attacking almost any simple optimization problem: 1. Obtain or manipulate a formula for the value-to-be-optimized so that there is only one free variable and everything else is a constant. 2. Solve for the value that makes the first derivative of this function zero. 3. Evaluate the function at that value.

 

[kuruman]

R is max at 45. That we have deduced. How can we show it here using this formula of R.
All I know is $cos\theta$ is max at zero.

That's not all you know. You also know that the symbol $t$ in the equation stands for the time of flight, and not just any time (that's why subscripts are useful). As stated earlier this time is $$t_{\!f}=\frac{2v_0 \sin\theta}{g}.$$

 

[rudransh verma]

1. Obtain or manipulate a formula for the value-to-be-optimized so that there is only one free variable and everything else is a constant. 2. Solve for the value that makes the first derivative of this function zero. 3. Evaluate the function at that value.

I don’t really get you. What value to be optimised? Are you talking about the derivative of $t=\frac{2{v_0}\sin\theta}{g}$
If you are, at $\cos 90$ derivative is zero.

 

[jbriggs444]

I don’t really get you. What value to be optimised? Are you talking about the derivative of $t=\frac{2{v_0}\sin\theta}{g}$
If you are, at $\cos 90$ derivative is zero.

We are trying to maximize R, not t.

 

[rudransh verma]

We are trying to maximize R, not t.

$\frac{dR}{d\theta}=-{v_0}\sin\theta t$

 

[jbriggs444]

$\frac{dR}{d\theta}=-{v_0}\sin\theta t$

No.

 

[rudransh verma]

No.

I don’t understand what you are saying in post#22

 

[jbriggs444]

I don’t understand what you are saying in post#22

What did I say there...

1. Obtain or manipulate a formula for the value-to-be-optimized so that there is only one free variable and everything else is a constant.

2. Solve for the value that makes the first derivative of this function zero.

3. Evaluate the function at that value.

You've skipped ahead to work in part 2 before you've completed part 1. You've also made the mistake of ignoring the fact that $t$ is a function of $\theta$.

So let us go back to the starting point:$$R=v_0 \cos \theta\ t$$We want a formula for the value-to-be-optimized ($R$) in which there is only one free variable and everything else is a constant.

$v_0$ is a constant for our purposes.
$\theta$ is a free variable.
But $t$ is neither a constant nor a free variable. It is defined as the time when the object hits the ground. Its value will depend on $\theta$. As has been pointed out more than once, it is given by:$$t=\frac{2v_0 \sin \theta}{g}$$
So we substitute this formula for $t$ into our formula for $R$ and obtain:$$R=v_0 \cos \theta\ \frac{2v_0 \sin \theta}{g}$$Now let us look at this formula. There is only one free variable: $\theta$. Everything else is a constant.

Now we are ready to proceed to step 2.

If we want to do it the hard way, we could keep our blinders on, apply the product rule for derivatives and get bogged down in some difficult algebra. Or we could get smart and apply the trig identity:$$\sin 2 \theta = 2 \sin \theta \cos \theta$$.
It is just a hop, a skip and a jump from there to concluding that the angle that maximizes $R$ is 45 degrees.

 

[rudransh verma]

@jbriggs444 I really appreciate all the hard work. But I am asking how $R={v_0}\cos\theta t$ tell $\theta=45$ will provide max R like the other formula for R.

 

[kuruman]

@jbriggs444 I really appreciate all the hard work. But I am asking how $R={v_0}\cos\theta t$ tell $\theta=45$ will provide max R like the other formula for R.

In your own words, please explain to us what the symbol $t$ stands for in the formula you quoted above.

 

[jbriggs444]

@jbriggs444 I really appreciate all the hard work. But I am asking how $R={v_0}\cos\theta t$ tell $\theta=45$ will provide max R like the other formula for R.

By itself, it does not.

By itself, eyes closed and looking at nothing but the formula on the page, the equation: $R={v_0}\cos \theta\ t$ appears to make R a function of three variables with no maximum.

It is when we open our eyes, use the information at hand and realize that $t$ depends on $\theta$ that we can see that for any given $v_0$ we have a function of one variable ($\theta$). This allows us to optimize for $R$ as a function of $\theta$.

This is not nearly as complicated as you are making it.

 

[rudransh verma]

In your own words, please explain to us what the symbol $t$ stands for in the formula you quoted above.

Time. I don’t know how t is a function of $\theta$. Is it because what you demonstrated in post#19
@kuruman did you assume in post #19 that let’s take both the eqns of R and equate it and we get t= something. When we did it with vertical eqn of motion we get the same t.
So because we get same t our assumption is right that both the eqns of R is same? Yes?

 

[haruspex]

Time. I don’t know how t is a function of $\theta$. Is it because what you demonstrated in post#19
@kuruman did you assume in post #19 that let’s take both the eqns of R and equate it and we get t= something. When we did it with vertical eqn of motion we get the same t.
So because we get same t our assumption is right that both the eqns of R is same? Yes?

You cannot presume to maximise R by setting theta to zero in $R=v_0\cos(\theta)t$. If you change theta you change t. In particular, if you set theta to zero the take off is horizontal, so you land immediately, making t zero, so R=0.
Using the 'function of' notation, we would write $R(\theta)=v_0\cos(\theta)t(\theta)$.
Differentiating, $\frac{dR(\theta)}{d\theta}=-v_0\sin(\theta)t(\theta)+v_0\cos(\theta)\frac{dt}{d\theta}$. To evaluate $\frac{dt}{d\theta}$ we need the formula for how t depends on theta.

 

[kuruman]

Time. I don’t know how t is a function of $\theta$. Is it because what you demonstrated in post#19
@kuruman did you assume in post #19 that let’s take both the eqns of R and equate it and we get t= something. When we did it with vertical eqn of motion we get the same t.
So because we get same t our assumption is right that both the eqns of R is same? Yes?

Yes. I think your confusion lies in your use of symbols. When you write $x=v_0 t \cos\theta$, the symbols in the equation are place holders. Symbol $x$ is the horizontal position at any time $t$ when the projection angle is $\theta$ and initial speed $v_0$. Symbol $R$ stands for the horizontal position at a specific time, namely the time when the projectile lands at the same height from which it was launched. That time is known as the time flight and is usually written as $t_{\!f}.$

So when you relplace "any horizontal position x" with the specific choice of $R$, namely the horizontal range, you have to replace $t$ on the right-hand of the equation with the specific time at which $x=R$. Thus you should write $R=v_0t_{\!f}\cos\theta$. Symbol $R$ now is the horizontal position at the specific time $t_{\!f}$ when the projectile lands. This specific time is related to the initial speed and launch angle in a manner that has been shown already.

As noted previously, subscripts are important; their omission often leads to confusion. It looks like you got stuck into thinking of $t$ in the equation $R=v_0t\cos\theta$ as "any time time $t$" when it is actually the specific time $t_{\!f}.$

 

[rudransh verma]

rudransh verma said:
I am questioning the existence of this formula. When you can do it via the obvious way by finding t why have we derived a new formula?

As I posted, we first have to find t without using θ=π/4. We can do that by writing the two usual equations:
y=vsin⁡(θ)t−12gt2, x=vcos⁡(θ)t
Substituting y=0 to find the time when it lands:
2vsin⁡(θ)t=gt2, 2vsin⁡(θ)=gt
Plugging that into the x equation:
x=vcos⁡(θ)2vsin⁡(θ)/g=v2sin⁡(2θ)/g
Now we can see x is maximised by θ=π/4.

So if I understood it correctly 1) we cannot find t because we don’t know at what $\theta$ t is such that R is max provided t is a function of $\theta$.
2) t is a function of $\theta$ shown via eqn $0={v_0}\sin\theta t-\frac12gt^2$

 

[rudransh verma]

It is just a hop, a skip and a jump from there to concluding that the angle that maximizes R is 45 degrees.

This is the standard crank-and-grind approach that you should try when attacking almost any simple optimization problem: 1. Obtain or manipulate a formula for the value-to-be-optimized so that there is only one free variable and everything else is a constant. 2. Solve for the value that makes the first derivative of this function zero. 3. Evaluate the function at that value.

$$R={v_0}\cos\theta t$$
Put the value of t we obtained from
$0={v_0}\sin\theta t-\frac12gt^2$
$$R=\frac{{v_0}^2\sin2\theta}{g}$$
Differtiating R with respect to $\theta$ and equating to zero we get $\cos2\theta=0$. $\theta=45$ . Putting in original eqn of R we get R maximised.

 

[haruspex]

The differentiation is unnecessary since we know the max value of sine is 1.

 

[rudransh verma]

The differentiation is unnecessary since we know the max value of sine is 1.

By the way what is the purpose of differentiating(I guess to find instantaneous rate) but why have we equated it to zero?
All we were doing was trying to maximise R.

 

[jbriggs444]

By the way why are we differentiating and why have we equated it to zero?

Recall our goal. We are trying to find the maximum range $R$ for a projectile launched at a known velocity $v_0$ at an unknown angle $\theta$ when the landing position is at the same elevation as the launching position. We are free to pick $\theta$ in order to maximize $R$.

We have come up with a formula that gives the achieved range, call it $R(\theta)$ for any given angle. That formula is:$$R(\theta)=\frac{{v_0}^2 \sin 2\theta}{g}$$This formula has been given many times already in this thread.

When one has an arbitrary function $f(x)$ defined for $x$ in some interval, there only a few places where a maximum or a minimum can be.

The maximum can be at one of the end points of the interval. For example, the function $f(x)=x$ on the closed interval [0,1] has its maximum at the right hand end point and its minimum at the left hand end point.

The maximum can be at a place where the function is discontinuous or where it is not differentiable. For example, the function $f(x) = |x|$ on the closed interval [-1,1] has its minimum at the bottom of the V that the graph traces out. The first derivative there is undefined.

The maximum can be at a place where the first derivative of the function is zero. For instance, the function $f(x)=\sqrt{1-x^2}$ on the closed interval [-1,1] has its maximum at the peak of its circular arc where the first derivative is zero.

One crank-and-grind approach to optimization problems is, accordingly, to solve for places where the first derivative is zero. The maximum may be at such a place.

As @haruspex points out, once one has a function of the form $$\text{some positive constant} \times \sin 2\theta$$there is no need to go the trouble of differentiating. It is clear by inspection where the maximum is to be found. [Still, nothing stops you from differentiating and solving $\text{some positive constant} \times \cos 2\theta = 0$ for $2\theta$. Easily enough done. This yields the correct result, as it must]

If no such obvious solution is to be found and if one is being careful [as one should be], one should check at the endpoints of the interval (e.g. at $\theta = 0 \text{ degrees}$ and $\theta = 90 \text{ degrees}$), at any possible points of discontinuity and at any points where the first derivative is zero or undefined to see at which point(s) the maximum (or minimum) is actually achieved.

 

[kuruman]

Here is another way to look at it. First rewrite the range $R$ in terms of the initial velocity components, $v_{0x}$ and $v_{0y}$ this way
$$R=\frac{v_0^2 \sin(2\theta)}{g}=\frac{v_0^2 (2 \sin\theta\cos\theta)}{g}=\frac{2(v_0 \sin\theta)(v_0\cos\theta)}{g}=\frac{2v_{0x}v_{0y}}{g}=\frac{2}{g}v_{0x}v_{0y}.$$Note that you can think of the range as proportional to the area of a rectangle of base $v_{0x}$ and height $v_{0y}$, the constant of proportionality being $\frac{2}{g}$. For an arbitrary choice of base and height you have two rectangles of the same area. For example, a projectile launched with $v_{0x}= 3~\text{m/s}$ and $v_{0y}= 4~\text{m/s}$, will have the same range (but different projection angle) as a projectile launched with $v_{0x}= 4~\text{m/s}$ and $v_{0y}= 3~\text{m/s}$. Because there are two different projection angles with the same range, the range cannot be maximum (cannot have the largest value) at these angles.

However, when the interchange of $v_{0x}$ and $v_{0y}$ results in the same projection angle $\theta$, you have a maximum^* because there is only one value for the range at that angle. This happens when $v_{0x}=v_{0y}$, i.e. the horizontal and vertical components of the initial velocity are equal and the projection angle is 45°.
__________________________________________________
^* It is a maximum because the range is zero when either one of the components is zero.

 

[jbriggs444]

Because there are two different projection angles with the same range, the range cannot be maximum (cannot have the largest value) at these angles.

I like arguments from symmetry. But there is a loophole in this one.

You've successfully proved that if the function has a unique global maximum then that maximum must be achieved when $v_{0x}= v_{0y}$. There is nothing that prevents a function $f(x)$ valid on the range [0, $\pi$] and satisfying the property that $f(x) = f(\pi-x)$ from having a non-unique global maximum in that range and then not achieving its global maximum at $x=\frac{\pi}{2}$

 

[kuruman]

I like arguments from symmetry. But there is a loophole in this one.

You've successfully proved that if the function has a unique global maximum then that maximum must be achieved when $v_{0x}= v_{0y}$. There is nothing that prevents a function $f(x)$ valid on the range [0, $\pi$] and satisfying the property that $f(x) = f(\pi-x)$ from having a non-unique global maximum in that range and then not achieving its global maximum at $x=\frac{\pi}{2}$

I am not sure I understand the loophole. If I choose a projection angle conventionally in the first quadrant of the unit circle such that $v_{0x}\neq v_{0y}$, then the product of the two, and hence the range, is doubly degenerate. This means that the range for this particular choice of angle cannot be an extremum. All angles in the first quadrant have a counterpart that results in the same range except the 45° angle which is its own counterpart. Thus, the range is at an extremum at 45°. The extremum is a maximum because the range is zero at $\theta =0$ and at $\theta = \pi/2$ and has positive values in-between.

 

[jbriggs444]

This means that the range for this particular choice of angle cannot be an extremum.

The symmetry argument guarantees that the range at some off-center angle cannot be greater than at all other angles. But that says nothing about whether it can be greater than or equal to the range at all other angles.

For instance, $f(x)=\sin 6x$ over the range $[0,\frac{\pi}{2}]$ has the symmetry properties that you are invoking. $f(x) = f(\pi-x)$. However, it has local and global maxima at $\frac{\pi}{12}$ and $\frac{5\pi}{12}$ and a global minimum at $\frac{\pi}{4}$. (15 degrees and 75 degrees for the twin maxima and 45 degrees for the global minimum)

 

[kuruman]

I think we are talking past each other. I am not invoking a general symmetry principle. I have an area which I will write as $A=v_{0x}\sqrt{v_0^2-v_{0x}^2}$ and maximize as $v_{0x}$ is varied from zero to $v_0$. It has only one global maximum when the two factors are equal.

 

[jbriggs444]

I think we are talking past each other. I am not invoking a general symmetry principle. I have an area which I will write as $A=v_{0x}\sqrt{v_0^2-v_{0x}^2}$ and maximize as $v_{0x}$ is varied from zero to $v_0$. It has only one global maximum when the two factors are equal.

So you are making use of the well known fact that the area of a rectangle of fixed perimeter is maximized when the rectangle is a square. Sure, that works.

 

[rudransh verma]

The maximum can be at one of the end points of the interval. For example, the function f(x)=x on the closed interval [0,1] has its maximum at the right hand end point and its minimum at the left hand end point.

I suppose the maximum of this function is at 1 as 1. Seems obvious.

The maximum can be at a place where the function is discontinuous or where it is not differentiable. For example, the function f(x)=|x| on the closed interval [-1,1] has its minimum at the bottom of the V that the graph traces out. The first derivative there is undefined.

The max here I suppose is at both 1 and -1.

The maximum can be at a place where the first derivative of the function is zero.

Is it a thumb rule?Is it a nature of graphs? I have not studied much about derivatives except how to do basic differentiation.

 

[haruspex]

Is it a thumb rule?Is it a nature of graphs?

Assuming the curve is smooth, if the gradient is >0 then the function is increasing, so hasn't reached the maximum yet; if the gradient is negative the function is decreasing, so has passed the maximum. To be at a local maximum (or minimum) the gradient must be zero.

 

[rudransh verma]

Assuming the curve is smooth, if the gradient is >0 then the function is increasing, so hasn't reached the maximum yet; if the gradient is negative the function is decreasing, so has passed the maximum. To be at a local maximum (or minimum) the gradient must be zero.

You mean when there is no rate + or - then the function is either at its maximum or minimum. The function is increasing/decreasing at an instant if the derivative is not equal to zero.

How do you know all this? Is it university grade stuff? All we are taught in 10+2 level is how to do differentiation and some differential equations.

 

[haruspex]

How do you know all this?

My post was intended as self evident.