Analyzing a function with its derivative | AP Calculus AB | Khan Academy
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- čas přidán 27. 08. 2024
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Taking the derivative of f(x)=x_-12x+2 and graphing the derivative, so we can tell when f is increasing or decreasing, and where is its relative extremum points. Created by Sal Khan.
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Many lives were saved with the help of this video
Oh man, someday when I get rich I will make a handsome donation to your channel cause you rock dude!!!!
math is so pretty, I wish it was nicer to me.
🤣same
Khan, saving kids doing calculus since 2013
WHAT THE FUNCTION
+abdilhakimable 5:30 :P
Lol thought the same
You can also plug in a lesser number such as f'(-3) if you wanna check the sign and then for f'(0) to check the sign and then a final f'(3) to check the sign and give you the graph without having to draw it. This logic works because remember if at f'(-2) is a zero we are checking for a sign change on both sides so if you do this method you are basically inputting values before and after (-2) to evaluate and you do the same for (2) of course.
5:31 >What the fu..nction would look like.XD
LOL i thought i was the only one.
*Critical point - where the derivative is equal to zero or does not exist/has a slope that equals zero.
-How to find: Find the derivative of the function and set equal to zero
-Each critical point can be labeled as a maximum or minimum
>Maximum point - critical point in which the derivative goes from positive to negative
>Minimum point - critical point in which the derivative goes from negative to positive
*If the left side of the point is going up, it is approaching positive infinity, therefore the derivative is positive
* If the right side of the point is going down, it is approaching negative infinity, therefore the derivative is negative
Graphing the derivative is sort of confusing. My teacher just taught us to make a chart and then just find where the derivative is positive and negative, making a local max/min. Just another method for someone who wants to try that.
after making the chart how did you find out the local max/min point
Magnesia _V i plugged in values to the right and left of each x root, then depending on whether that y outcomes were positive or negative, I could see if the graph had a change and therefore see if there was a local max / min (for ex.: if your root x was 0... if y outcomes to the left of 0 were negative, and y outcomes to the right of 0 were positive, that means there was a switch from negative to positive and therefore a local min. I’m sorry if that explanation wasn’t the best I’m just a student hope that helped tho!
You do know if you want an easier way to find maximum and minimum point, you can just differentiate it twice
That's just a working procedure!! You Need to understand the motive behind that!!
yeah but now you understand why we derivative the derivative !
The way you emphasized "What the f.." startled me. 5:31
x^2 is parabolic which have minimum at [0,0], when you substract 12, whole function goes down by twelf.
SAL I LOVE YOU FOR EXPLAINING THIS. MY FRUSTRATION AND SUFFERING HAS FINALLY ENDED.
I wish this was around when I was in high school. Thanks for continuing my education!
ur explanation is too good....u make people visualise..every thing thanx man.....
Between -2 and 2 how do you decide which is a maximum and when it is a minimum? Negative values are always maximum values and positive values are always negative values?
...Minimum...
Nice explanation
Love from India
This is such a great explanation. I wish my math teacher was like you.
Your voice sounda like music...no rather than a symhony..thanks a lot
Sal, In the Greek Open Univercity EAP, we also do the so called infumum and supremum of function sequences. Make some videos about that to when you can.
I have a question. As you've mentioned for a function to have a max / min value (i.e. for a function to have a critical point), f`(x) should be 0 and the value of x, thus obtained, +2 and -2 are considered as maximum / mimimum. This is where I am confused. Why have we considered that point as max/min value of the function? How do we confirm this, that it is a critical point not an inflexion point. I mean even inflexion point have f`(x) = 0 but there are no maximum / minimum value.
May be you can try hard examples. We want some insane problems. PLS
this guy is so cute the way he repeats the words is too cute
For finding Max and Min, you checked the slope changing from positive to negative. Why did you go right to left? if we check from left to right, f(2) becomes max.
you count the numbers from -infinity to +infinity and that's what is represented in the number line. hence he moved from -ve to +ve as per the numbers represented on x-axis.
Love your videos! Such a great supplement for my classes.
Excellent job. Thank you so much!
Thanks a lot
life saving videos indeed
Another Great Video..!
I have a similar problem to this one in my Stewart 8th ed text, its num 4.1 ex 50. F(x) = x^3-6x^3+5...[-3,5]...I'm stuck on the part where you find the derivative and set to zero. No problem, but the solutions manual has a different answer for that. Unlike what you did which was take the square of 4 and come up with critical points (+ or - 2)...This manual factors out like this...3x(x-4) and ends up with the critical points (0,4). Please clarify this. Thank you
good job Sal
I'm at 4th year level civil engineering and I dont rven understand any of these. What the function am I doing???
Kirby Calitis how the did you get it at the first place lol
@@abdib5446 😂😂😂😂😂
5:31 WHAT THE FUNCTION?
BRILLIANT KHAN
this guy has saved me too many times to count
thanks
what if we have an absolute value function?
how to solve y=3x^2-6x-1 this?
learn dear from these vedoes
Ur my live savior.... I needed help badly
why f'(x) is draw like this ?
didn't undersatnd how u ound out the val was max
Why the screen be green
how is it that f'(-2)=3(-2²)-12 equal to 0.... don't know how u arrived at that zero cuz am getting -24
The difference between "negative two, squared" (+4) and "negative, two squared" (-4) ;)
why graphs of f'(x) looks like this?
its a primitive graph of a parabola
If the derivative curve [y= f' (x) ] has a negative slope at/ beyond a point, it means that the actual curve [y= f (x) ] has a positive slope corresponding to that point/ beyond it and vice versa.
Correct me if I am wrong! :D
5:31
😂😂
Im too stupid to understand
🥰🥰
the only f(x) I want to care about is the girl group
Woot! First Comment!! Go Sal!