finding local maxima

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chillintoucan28

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I have the equation x^3-4x, I need to fing the coordinates of any local maxima of the derivative of that function (also think about when the derivative is positive and when it is negative)

Thanks!
 
Hi Chill'n:

Will you please post your work?

Do you know how to write the first derirvative?

Do you understand the phrase, "local maxima"?

I'm motivated to ask because your recent discussion about using a number line to solve inequalities actually turned out to be something else entirely.

Is this post about finding local maxima another experiment? If you are experimenting, then that's GREAT! I like to see students of math experimenting. However, if I'm involved, then I also like to know what's actually taking place.

Either way, experiement or not, please show some work for whatever it is that you are thinking. That way, we'll all be on the same page.

chillintoucan28 said:
I have the equation x^3-4x, ...
Click to expand...

From reading your other posts, I think that you've reached the point where one is careful about terminology.

x^3 - 4x is not an equation. All equations contain an equal sign. x^3 - 4x is an expression.

Cheers,

~ Mark
 
chillintoucan28 said:
i actually do not know what the phrase local maxima means... could you help me out
Click to expand...
Hmm... that's awkward....

They would have covered max/min stuff, along with derivatives, back in Calc I. You appear now to be in Calc II, beginning integration by relating integrals to derivatives. If you're not familiar with derivatives or any of that Calc-I stuff, your apparent utter confusion in Calc II certainly makes more sense. :shock:

You might want to have a serious talk with your academic advisor. I would not expect any student, no matter how brilliant, to succeed in Integral Calculus without a thorough understanding of Differential Calculus!

Good luck! :D

Eliz.
 
Hi Chilli'n:

You betcha.

Maxima is plural for maximum. Minima is plural for minimum.

Minimum and maximum are the points where a function achieves its smallest and greatest value, respectively.

On the function's graph, the minimum is the lowest point, and the maximum is the highest point.

Of course, a function may increase and decrease multiple times over the course of its domain. Each time the graph changes direction VERTICALLY, we end up with either a "peak" or a "valley".

If we focus on just one section of the number line (i.e., if we constrain the domain to a particular interval of x, then we're considering only ONE PART of the function's graph. The highest and lowest point(s) within this restricted domain are called the LOCAL maximum and minimum, respectively.

Please stand by ... dogs creating havoc downstairs ... be right back ...
 
Yeah, surprisingly, I am going into AP Calculus next year as an 11th grader, I have only taken pre-calc and i am doing problems to rpepare my self for next year and to be ahead of the class...

um, so then in finding the local maxima, why would I have to consider when the derivative is positive and negative? how can a derivative be both positive and negative?
 
chillintoucan28 said:
... in finding the local maxima, why would I have to consider when the derivative is positive and negative? how can a derivative be both positive and negative?
Click to expand...

At any given value of x, the derivative can have only one value. So, if you restrict yourself to looking at the function for only ONE value of x, then you are correct in saying that the derivative at that point can not be both positive and negative.

However, we don't examine functions at one point only; we examine functions' behavior over INTERVALS.

Within any given interval, a function's behavior often changes. Sometimes, it is increasing. Sometimes, it is decreasing. Sometimes, it is constant (i.e., not changing).

Consider the graph of a sine wave.

It has many "peaks" and "valleys" over the interval from x = -2*Pi to x = 2*Pi.

Here, I have constrained the domain to [-2 Pi, 2 Pi].

If we wanted to find the local maxima for f(x) = sin(x) in this domain, we would report the coordinates for each point on the graph where f(x) = 1.

The first derivative of the Sine function has different maxima because that is a different function.

If you're trying to learn introductory derivative calculus before your class starts, then I suggest you complete multiple Internet searches and devour as much info as you have time for. By examining multiple sources, you get to see the same info presented in many different ways.

~ Mark :)
 
ok so could you tell me as an example the coordinates for the local maxima of the derivative ox x^3-4x

look school has not started, its till summer... im not cheating or anything, this is for my on personal benefit. Msot kids are goofing around an playing right now, but I am wokring on math. Can you please do this for me...
 
chillintoucan28 said:
... im not cheating or anything, ...
Click to expand...

Nobody accused you of cheating.

The topic of global and local behavior of functions is a pre-calculus topic.

In other words, we can talk about maxima and minima in any class that understands the concept of functions.

So, I'm going to provide you with the first derivative function for the original function you posted. Then, you can examine the graph yourself.

f '(x) = 3x^2 - 4

Do you recognize what type of graph f '(x) has?

Those of us at this board could type up an entire lesson with multiple examples to answer your original question, but that is not the purpose of this board. So, we need to take things one step at a time.

Again, I think it's great that you're applying yourself when your friends are out playing. You will go far in life.

~ Mark
 
Hi Chill'n:

I reread the entire thread, and I found a concern.

... coordinates for the local maxima of the derivative ox x^3-4x ...
... coordinates of any local maxima of the derivative ...
... in finding the local maxima, why would I have to consider when the derivative is positive and negative? ...
Click to expand...

f(x) = x^3 - 4x

The first derivative function for f is:

f '(x) = 3x^2 - 4

BOTH f(x) and f '(x) are functions.

BOTH have their own local maxima.

You asked for the local maxima of: f '(x).

It seems like maybe you are thinking about USING the function f '(x) to find the local maxima FOR f(x).

If this is true, then you have not typed what you want.

Please clarify for me.

Do you want to find local maxima for x^3 - 4x instead?

~ Mark :)

 
Yes, sorry about that, we want to find the coordinates local maxima of x^3-4x but again we have to keep in mind when the derivative is positive and when it is negative
 
Question: How does the first derivative of x^3-4x being positive or negative have anything to do with finding local maxima for x^3-4x?

Answer: The sign of the first derivative tells you whether or not x^3-4x is decreasing or increasing. (If the first derivative is zero, then that means x^3-4x is neither increasing nor decreasing.) When x^3-4x changes direction, it EITHER goes from increasing to decreasing OR from decreasing to increasing. This means that the first derivative must change sign at any point where x^3-4x changes direction. If you calculate the value(s) of x where the first derivative changes sign, then you will get the values of x where the graph of x^3-4x changes direction. These points are the "peaks" and "valleys" on the graph of x^3-4x. Analyze (investigate) what is happening in the vicinity of each value of x you discover from the first derivative to see which point is a maximum.
 
chillintoucan28 said:
do you know the actual answer, like coordinates of the local maxima?
Click to expand...

No, I do not know the actual coordinates*.

Since the term 'maxima' is plural, I do not even know what interval of the domain you are considering to be local. I could guess that local means "in the vicinity of every peak" AND "in the vicinity of every valley", but I'm not sure whether or not you agree.

If you meant to type "the local maximum", then I understand that you want the coordinates of a single point.

I do not know the actual coordinates of this point, but I know how to determine them for what I consider local.

I would calculate the derivative function and see that it is defined by 3x^2 - 4.

I would realize that I need to know for which values of x the expression 3x^2-4 equals zero because that will tell me where the value of 3x^2-4 changes sign, which, in turn, tells me where the peaks and valleys are located on the graph of x^3-4x.

So, I would need to examine the behavior of x^3-4x AT THOSE VALUES of x to see what's happening.

3x^2 - 4 = 0

If you know how to solve this equation for x, then you will have the values of x where the behavior of function f changes direction.

f(x) = x^3 - 4x

If you LOOK at the graph of f(x) AT THOSE VALUES of x, then you will see these behavioral changes as a "peak" and a "valley".

Then, it should be clear enough which point is a maximum and which point is a minimum. The maximum point is called a local maximum when you consider only a section of the domain. Namely, a short section that contains only the peak of the graph.

You will then know which of the two values that you got from solving the quadradic equation above is the x-coordinate of the local maximum. If this value turns out to be, oh, say -4, then the y-coordinate is f(-4). If it turns out to be -1.265, then the y-coordinate is f(-1.265).

In other words, the *actual coordinates are (X, f(X) ), where X is one of the two numbers discovered from solving 3x^2-4=0.

Please feel free to continue to ask specific questions. ~ Mark :)
 
chillintoucan28 said:
do you know the actual answer, like coordinates of the local maxima?
Click to expand...
You say that you are trying to learn. Then why not try to use the mini-lessons that have been attempted here, along with what you've learned in past courses and what you can see in your current text, and try to do the exercise yourself? The mere numerical answer doesn't "explain" or "teach" anything, so of course that can't be what you really want.

So please study all of the various bits of information with which you have been provided, and attempt the exercise yourself. Then, if you get stuck, you will be able to reply with a clear listing of your work and your reasoning. Thank you! :D

Eliz.
 
ok Mark, so using the quadratic formula I was able to solve the equation

My answers were negative (square root of 48)/12

and positive (square root of 48)/12

that doesnt look right, is it correct?
 
chillintoucan28 said:
... using the quadratic formula I was able to solve the equation 3x^2 - 4 = 0

My answers are negative (square root of 48)/12

and positive (square root of 48)/12

that doesnt look right, is it correct?
Click to expand...

Nope, those are not the correct values for x.

It's curious to me why you would post, "doesn't look right", since you don't know at this point what the values are for sure.

What do you EXPECT these values to "look like"? What makes you SUSPECT that they are not correct?

Whatever your reason(s), your speculation turns out to be valid ... they're not correct.

We can verify their accuracy by using them to evaluate the original function.

f '(x) = 3x^2 - 4

f '(?48/12) = -3

f '(-?48/12) = -3

We want f '(x) = 0

You did not show any work, so I don't know how you came up with the wrong values for x. :!:

You wrote that you used the quadradic formula to solve f '(x) = 0. That's certainly an option; however, may I suggest another way? Start by adding 4 to both sides ...

Cheers,

~ Mark :)

PS: It would be quite sloppy if a calculus student were to report ?48/12 for anything because this expression can obviously be REDUCED.
 
well, the reason I can SUSPECT it, is because I am looking a graph...

so we are evaluating 3x^2-4=0

so we get 3x^2=4

then we get 3x=2 by taking the square of both sides, so x=(2/3), right??
 
chillintoucan28 said:
well, the reason I can SUSPECT it, is because I am looking a graph...

so we are evaluating 3x^2-4=0

so we get 3x^2=4

then we get 3x=2 (Incorrect) by taking the square of both sides, so x=(2/3), right??
Click to expand...

\(\displaystyle 3x^2 \, = \, 4\)

\(\displaystyle x^2 \, = \, \frac{4}{3}\)

\(\displaystyle x \, = \, \pm\sqrt{\frac{4}{3}}\)
 
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