Smoothing a Piece-wise Function | MIT 18.01SC Single Variable Calculus, Fall 2010
ฝัง
- เผยแพร่เมื่อ 6 ม.ค. 2011
- Smoothing a Piece-wise Function
Instructor: Christine Breiner
View the complete course: ocw.mit.edu/18-01SCF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
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So this is what a good education looks like
This lady is perfect. This is definitely going to help me pass my calculus exam. Thank you for posting this! I owe my life to you, haha.
did you pass? what are you doing today
Wow, wow, you explain soooooooo clearly, that so easy to understand. I'm very glad to find you. Thank you
This is another amazing lecture on Smoothing a Piece-wise Functions by Professor Christine Breiner.
Amazing?
thanks a lot MIT, i understood everything, greetings from Colombia :D
Thanks for the presentation.
Thank the speaker clearly stated what's continuous and differentiate properties.
@Simon Hoogendal: The derivatives of left and right are never going to be equal in terms of their actual function definitions. The left side of the piecewise function is a linear function with a constant slope, and the right is a parabola with a variable slope. No matter how you define the left side of the function, this will always be the case since the problem tells us that the function must be linear at x 1 of the derivative function of each must be equal. If they aren't, it means that the slope of the tangent line at x=1 is different in each function. So again, the derivative functions of each part of the piecewise function will never be equal in definition, but they must each give the same output when x=1.
Thank you so much.
I LOVE MIT OCW!
This is only possible line ax+b. b/c exactly at (1,2) there's a tangent line that has to coincide with the line ax+b in order to be smooth, however you can skip this step by just saying f(x) is riemann integrable. As a side note: all the solutions for ax+b, (a,b) is isomorphic to R^2.
In the second part, you impose that the function be *continuously differentiable at x=1, which is technically stronger than differentiability. For some functions, f'(a) might exist even if the limit of f'(x) as x goes to a does not.
thank you...
Gotta go to MIT to see the artwork.
someone help me with the first part.does x^2+1 = ax+b for the function to be continuous?
Jake Goykia the definition of continuous is limit of f(x) approach to 1 from left and right must be the same, hence a + b must equal to 2.
but didn't you use a different definition for the left/right hand limit? How is that right.
If a=x and b = 1, so the second function was y=(x^2)+1, then wouldn't the resulting function have a derivative, because the combination would just create the function y=(x^2)+1 ?
I get it !!! I FINALLY GET IT ! WOOOHOOO
Why do we need a limit of left and right derivative to be equal, not the left and right derivative itself, for a function to be differentiable?
I have the same question.
good job
lastima esta en ingles, podrian poner subtitulos en traduccion en español,ojala puedan sino ni ablar, gracias_Mis-Christine Breiner
Super
Genius
but even if b not = 0 the derivative of ax + b will be 2 if a = 2 ?
soumya sarkar b must be zero because the condition of a+b=0 dude.. there is two condition must be fulfil, differentiable and continuous.
sorry... i mean from a+b=2 😅
Your right except for the fact that F is differentiable at all reals greater 1. Because if x>1, I can choose h>0 small enough that [F(x+h)-F(x)]/h>0.
I was really hoping the cameraman would pan out to show the artwork at the end :(
If you missed it look at 7:13
7:13
1:51 SLOW MOTIONS HUE HUE HUE HUE
Amejing platform
Ojala Christine haga cursos de calculus,, la entiendo muy bien
too bad the function is not "smooth"
Nice