Retarded Maths Question(derivatives)

[7254278]

Hey all,
Tomorrow I got a pretty easy math exam. Ive studied more or less enough but there are 2 thing I just dont know how to do. One thing is find derivatives in simple formulas like this one :2X^3-5X^2+7X-1. Can some one simpley explain to me how to find derivates and another thing I dont know how to do is find the maximum and minimum points(like she gives us the formula for a graph and we have to find the maximum and minimum points without using a calculator, I think it involves derivates aswell) can someone also explain to me how to do this. Keep in mind this is a highschool test so all the formulas and stuff we have to solve are pretty simple and straight forward so there is no ned to go too in depth.
Sucks that I get As in english and computers but I get Ds in lower level maths .

 

[atszyman]

Hey all,
Tomorrow I got a pretty easy math exam. Ive studied more or less enough but there are 2 thing I just dont know how to do. One thing is find derivatives in simple formulas like this one :2X^3-5X^2+7X-1. Can some one simpley explain to me how to find derivates and another thing I dont know how to do is find the maximum and minimum points(like she gives us the formula for a graph and we have to find the maximum and minimum points without using a calculator, I think it involves derivates aswell) can someone also explain to me how to do this. Keep in mind this is a highschool test so all the formulas and stuff we have to solve are pretty simple and straight forward so there is no ned to go too in depth.
Sucks that I get As in english and computers but I get Ds in lower level maths .

The derivative of ax^y = yax^(y-1) if you apply this formula to every term through the above expression you will get the derivative of the formula. 2x^3 => 6x^2, -5x^2 => -10x^1 or -10x, 7x => 7x^0 or 7, -1 or -1x^0 => 0. So the derivative is 6x^2-10x+7.

To find the maxima/minima find where the derivative is equal to 0. This corresponds to the point on the line where it has zero slope. There are advanced methods to finding out if it is a maximum or minimum by looking at higher order derivatives but it sounds like you won't have to do that if you have the graph in front of you.

 

[floriflee]

Ah, calculus... was so fun 10 years ago. Let's see what I remember....

Derivatives.... an easy way to figure it out (assuming I remember correctly, and I'm sure someone here can verify)...

Assume you have the following:

Ax^b + Bx^c + Cx^d +D

the formula for the derivative is something like this:

(A*b)x^(b-1) + (B*c)x^(c-1) + (C*d)x^(d-1)

So your example:

2X^3-5X^2+7X-1.

would be

6X^2 - 5X +7

 

[Oryan]

So your example:

2X^3-5X^2+7X-1.

would be

6X^2 - 5X +7

You forgot the power on the second term. The answer should be 6X^2 - 10X + 7.

 

[plinden]

These days, does anyone actually learn what derivatives represent, rather than a rule of thumb on how to derive them from the initial function?

For instance atzsyman says that the min/max of the function is where the derivative evaluates to 0 ... so, do you know why it evaluates to 0 at the min and max?

I think getting a feel for this would help you understand calculus in the future. My maths (yes, maths, not math) teacher taught us this from first principles, and I never forgot it.

Ah, maths, I miss it. I did engineering maths as an elective all the way through four years of college.

 

[floriflee]

You forgot the power on the second term. The answer should be 6X^2 - 10X + 7.

Oops.. done in haste... can't even follow my own formula for solving.

 

[whocares]

For instance atzsyman says that the min/max of the function is where the derivative evaluates to 0 ... so, do you know why it evaluates to 0 at the min and max?

And atzyman is [at least partially] wrong. In mathematical terms:

IF you're at a min or max, then the derivative is equal to zero.


A derivative equal to 0 means you have reached a local min or max.

To get what I mean, try plotting
y=x^3 - 4x
This function's derivative has 2 x values where it is equal to 0, but these x values have nothing to do with the min of max of y.

--------------------

How to find the min-max of a function, or how to study a function y=f(x)

1. Study the limits of the function: what values will y reach when x tends towards infinity?
2. Calculate the functions derivative, f'(x)
3. Determine for what x (there can be several), f'(x)=0. This will give you local highs and lows.
4. Determine the sign of f'(x) in between the previous x values you find (this will tell you if f(x) is increasing or decreasing over that interval).
5. Calculate y for these same x values. These y values will be the local mins as max.
6. As you have studied the function from -infinity to +infinity, you now know the absolute max and mins.

I'll do an example in my following post

 

[7254278]

Ok derivitives now I get but I dont get this min/max stuff, how would i find the min/max on the formula i gave u guys above?

 

[atszyman]

And atzyman is [at least partially] wrong. In mathematical terms:
...
A derivative equal to 0 means you have reached a local min or max.
...

To be fair I never specified anywhere whether the minimum or maximum found where dy/dx = 0 to be local, global, or otherwise. Since the original poster indicated that things would not be that difficult I was under the impression that the functions used by their instructor would have an obvious local/global max/min that would be the object of the question. That and if I get too sidetracked I will end up in math that nobody wants to deal with (the joys of currently earning my Master's degree in Electrical Engineering).

 

[whocares]

Polynome study: example

Let's define a function f such as:

f(x) = 3x^3 - 9x

This function is defined from -infinity to +infinity, ie all 'Real' values of x will have an f(x) value.

1. Let's see what happens at the functions limits.
When x approches -infinity, f(x) also approaches -infinity. I'm unsure if you've studied limits yet, but this is basically saying that for very small values of x, the smaller x gets, the smaller f(x) gets.
Likewise, when x approches +infinity, so does f(x).
We now have absolute mins a maxs for the function at the function's limits.

2. Let's look a the derivative, f'(x):

f'(x) = 9x^2 - 9

This f'(x) has 2 roots, ie 2 values of x where f'(x)=0. These are 1 and -1. Hence we can re-wite f'(x) as:

f'(x) = (3x - 3)(3x + 3)

if you don't trust me, try replacing x with 1 in the first brackets.

We now know that f(x) has local min or max at x=-1, and the same at x=1

3. Let's look at the sign of the derivative :

if x<-1, then (3x - 3) < 0 AND (3x + 3) < 0 ==> f'(x) > 0
if x>1, then (3x - 3) > 0 AND (3x + 3) > 0 ==> f'(x) > 0
but if -1<x<1, (3x - 3) < 0 AND (3x + 3) > 0 ==> f'(x) < 0

So to sum things up:
if x<-1, f'(x) > 0, f(x) is increasing
if x>1, f'(x) > 0, f(x) is decreasing
if -1<x<1, f'(x) < 0, f(x) is increasing

As f(x) is increasing before x=-1, but decreasing after x=-1, x=-1 marks a local maximum for f(x). Like wise, x=1 is a local minimum for f(x).

4. So to sum everything up:

From -infinity to -1, f(x) is increasing to a maximum value of 6
From -1 to +1, f(x) is decreasing to a minimum value of -6
From +1 to +infinity, f(x) is increasing away from -6.

-----

I hope this is clear enough and helps you. You should plot out f(x) and f'(x) to follow the reasonning.

 

[dukebound85]

To get the min max of a function,

Just take the derivative of the derivative you got. Say your inital equation is y= 2x^2+3x. dy/dx=4x+3. To get the point at the min or max, set this to 0 and solve for x. in this case dy/dx=0 makes x = -.75. This is the point at which the slope is 0 making it a min or max. To determine if it is a min or max all you do is take the derivative again (ie dy^2/d^2x). When doing so, dy^2/d^2x=4. Since this value is positive, the function is a min at x=-.75. If the second derivative happened to be positive, that would make it a max. If you have multiple mins or max's, you would get multiple x values in the dy/dx=o step. Just a matter of evaluating them and seening what the corresponding y values are. Hope this helps

 

[whocares]

Polynome study: exampe (cont'd)

5. roots of f(x):

Now that we know the "shape" of f(x), we know how many roots it has: 3; ie there are 3 values of x for which f(x)=0.
One is between -1 and 1 (f(x) is decreasing from 6 to -6, so must be equal to 0 somewhere). One is between -infinity and -1 (for the same reasons), and the last between +1 and +infinity.

One is pretty easy, it's x=0:

f(x) = 3x^3 - 9x
<=> f(x) = 3x (x^2 - 3)

There are formulae for easily solving second order polynomes (that's how I did step 2.). So (trust me):

f(x) = 3x (x - √3) (x + √3)

6. The end.

We now know everything there is to know about f:
* its "shape"
* its absolute mins and maxs, as well as its local mins and maxs
* where it's equal to 0

----

Application to all polynomes.

Of course this works for all polynomes. If the order of the polynome is greater than 3, you just have to do more steps. For example, a 4th order polynome, you can first study its derivative (a 3rd order polynome as above).

Of course you can use the same general method on other, more complex, types of functions.

 

[plinden]

Ok derivitives now I get but I dont get this min/max stuff, how would i find the min/max on the formula i gave u guys above?

To keep it simple, the derivative of a function y=f(x) specifies its gradient at any x.

Your function y = 2x^3-5x^2+7x-1 has the derivative dy/dx = 6x^2 - 10x + 7

So, for example:
x = 0, y = -1, dx/dy = 7

Now, if you plotted a function you may see points where the gradient changes from positive to negative or from negative to positive. At these points the gradient is 0, ie a minimum or maximum (or has been pointed out, local max or min). A maximum is where the gradient changes from positive to negative and vice versa for the minimum. (Plot this out to get a feel for it)

So for your function, the local minimum or maximum is where dy/dx = 0, or
6x^2 - 10x + 7 = 0

Using x=(-b+/-sqrt(b^2-4ac))/2a, this provides the result (10 +/- sqrt(100 - 168))/12

Unfortunately, sqrt(-68) is imaginary, so your function has no local minimum/maximum.

 

[Plymouthbreezer]

It's like personalized homework help!

 

[7254278]

It's like personalized homework help!

Hehehehe, what can macrumors NOT do?
Well I took the exam there were 4 questions.
2 on derivatives which I am 90% I got right(THANK YOU MR, plinden, whocares, atszyman!)
1 on finding the max/min which I am 50% sure I aced, I atleast got partial points(thanks dukebound85).
1 question having something to do with tangents which I unfortunatley skipped. I am hoping to get atleast 50% on this test, I only wanna pass maths which for me is almost impossible!
Once again thank you MR!

 

[bemylover]

How to find the min-max of a function, or how to study a function y=f(x)

1. Study the limits of the function: what values will y reach when x tends towards infinity?
2. Calculate the functions derivative, f'(x)
3. Determine for what x (there can be several), f'(x)=0. This will give you local highs and lows.
4. Determine the sign of f'(x) in between the previous x values you find (this will tell you if f(x) is increasing or decreasing over that interval).
5. Calculate y for these same x values. These y values will be the local mins as max.
6. As you have studied the function from -infinity to +infinity, you now know the absolute max and mins.

Great job whocares That is actually everything you can do to study a function.

I will just add that there's another method to find out whether f'(x)=0 is min or max. If f''(x) (second derivative) at that point is positive, you have minimum and if it's negative, you have maximum.

cheers