I appreciate you Mr Newton. Ur creativity and deep understanding of Maths is really Amazing. I watched most of Ur video and I got unforgettable concepts.
I've just found your channel and I love it! You bring such clarity to a subject that, while I love it, can be very confusing. Thank you for all you're doing!
@@sloosh2188 not much honestly. We went over piecewise graphs for limits which was confusing when I had never seen piecewise functions before. But once we got to differentiation and beyond we abandoned piecewise functions.
@@JourneyThroughMath That's all my calc 1 class did with them as well, interesting to me that you dont remember much from it since i feel like i remember every god damn thing from those lessons lol. while theyre not practical in real life (really at all) theyre a great tool for actually understanding the definition of differentiability/ continuity, but if that was your first introduction to them i can totally see how it didnt really stick with you. What math are you taking now bro? im in Diff EQ and lin alg currently.
@@sloosh2188 Im out of college. Im a math teacher. I like to stay up to date because I enjoy math but its good to keep my options open for what I might one day teach
The equation of the tangent at the graphic representation of f: x ---> x^2 + x at x = -3 is y = f'(-3).(x +3) + f(-3) = -5.(x +3) + 6 = -5.x -9 So the function will be C1 on -3 (and on R) if a.x + b is identical to -5.x -9, so a = -5 and b = -9 (it will not be C2)
Did you forget an “or equal to” in the first part of the piecewise function? Correct me if I’m wrong (I’m just an average high school math student) but if x = -3 you can only use the second part of the piecewise function (the line with unknown coefficients and constants is only used for values less than -3) so I don’t understand how the derivatives have to be equal at x=-3.
Before watching the video: Only point where this function couldn't be differentiable is x=-3. In order for it to be differentiable there it first needs to be continuous, so lim x-->-3 of f(x) must exist or lim x-->-3 from the right and left are equal. From the left we see this is lim x-->-3 (Ax+B)= -3A+B and from the right lim x-->-3 (x^2+x)= (-3)^2+(-3)=9-3=6, setting them equal gives 1st equation is -3A+B=6. Next the derivatives of both piecewise branches should be equal at x=-3 to have a defined derivative so (Ax+B)'=(x^2+x)' at x=-3 or A=2x+1 at x=-3 so A=2(-3)+1=-6+1=-5. Plugging this back into the first equation yields -3(-5)+B=6 or 15+B=6 so B=-9. Our solution is thus A=-5 and B=-9.
That was not a correct problem description. What you SHOWED in the video was DOUBLE differentiability. Differentiability in the SECOND derivative. What you ASKED for in the presentation at the beginning was SIMPLE differentiability. Differentiability in the FIRST derivative. Since for the simple differentiability, the curves have just to match at the overturn point, you may chose one of the parameters freely.
@@priyansharma1512 That was what he showed in the video. But what he ASKED for in the thumbnail of the video - thus in the problem presentation - was ONLY differentiablity. ... Well: you could argue that there was the term "continuity" in the title. But it was not in the title as a part of the requirements, only as a description of the aspects which he will be talking about.
I really want to know how you want the problem be described. So it matches what I did in the video. We Never Stop Learning. I'll change the description if it makes sense.
@@PrimeNewtons Maybe in the thumbnail write : "f ' (x) differentiable" (the derivative has to be differentiable, not the original f) ...wait a moment: Could it be that the apostroph is already there but just not visible because the thumbnail being too small? ...No: Two videos earlier the apostroph is clearly visible... I really think it's missing in this thumbnail.
“Piecewise Smooth” was going to be my rap name, but that career never took off.
I appreciate you Mr Newton. Ur creativity and deep understanding of Maths is really Amazing. I watched most of Ur video and I got unforgettable concepts.
I've just found your channel and I love it! You bring such clarity to a subject that, while I love it, can be very confusing. Thank you for all you're doing!
"A = -5. 'No way!' "
He's teaching the concepts, but he's still astonished by the answer. You gotta love the enthusiasm. 😇
i have known how to solve these kind of problems but thanks to you i got in depth meaning and knowledge :)
I love your energy so much, you seem to have so much passion for math ❤
Great follow up to your other video about showing a continuous function.
,حقا أنت رائع
Beutiful!!! Thanks Newtons
This is the perfect type of problem. I have never seen this type of problem before but I know I have all the tools to tackle it
Did your calculus class not talk about piecewise? Lol
@@sloosh2188 not much honestly. We went over piecewise graphs for limits which was confusing when I had never seen piecewise functions before. But once we got to differentiation and beyond we abandoned piecewise functions.
@@JourneyThroughMath That's all my calc 1 class did with them as well, interesting to me that you dont remember much from it since i feel like i remember every god damn thing from those lessons lol. while theyre not practical in real life (really at all) theyre a great tool for actually understanding the definition of differentiability/ continuity, but if that was your first introduction to them i can totally see how it didnt really stick with you. What math are you taking now bro? im in Diff EQ and lin alg currently.
@@JourneyThroughMathno fucking way, i just checked your channel and i wouldnt be surprised if youre literally taking the exact same courses i am rn
@@sloosh2188 Im out of college. Im a math teacher. I like to stay up to date because I enjoy math but its good to keep my options open for what I might one day teach
I being in a low class am able to study these due to this channel❤
Since differentiability implies continuity, it might be a good idea to work the derivative part before you work the limit part .
I agree.
Thank you so much sir
The equation of the tangent at the graphic representation of f: x ---> x^2 + x at x = -3 is y = f'(-3).(x +3) + f(-3) = -5.(x +3) + 6 = -5.x -9
So the function will be C1 on -3 (and on R) if a.x + b is identical to -5.x -9, so a = -5 and b = -9 (it will not be C2)
Did you forget an “or equal to” in the first part of the piecewise function? Correct me if I’m wrong (I’m just an average high school math student) but if x = -3 you can only use the second part of the piecewise function (the line with unknown coefficients and constants is only used for values less than -3) so I don’t understand how the derivatives have to be equal at x=-3.
Before watching the video: Only point where this function couldn't be differentiable is x=-3. In order for it to be differentiable there it first needs to be continuous, so lim x-->-3 of f(x) must exist or lim x-->-3 from the right and left are equal. From the left we see this is lim x-->-3 (Ax+B)= -3A+B and from the right lim x-->-3 (x^2+x)= (-3)^2+(-3)=9-3=6, setting them equal gives 1st equation is -3A+B=6. Next the derivatives of both piecewise branches should be equal at x=-3 to have a defined derivative so (Ax+B)'=(x^2+x)' at x=-3 or A=2x+1 at x=-3 so A=2(-3)+1=-6+1=-5. Plugging this back into the first equation yields -3(-5)+B=6 or 15+B=6 so B=-9. Our solution is thus A=-5 and B=-9.
Can you do physics
That was not a correct problem description.
What you SHOWED in the video was DOUBLE differentiability. Differentiability in the SECOND derivative.
What you ASKED for in the presentation at the beginning was SIMPLE differentiability. Differentiability in the FIRST derivative.
Since for the simple differentiability, the curves have just to match at the overturn point, you may chose one of the parameters freely.
No the 1st part was the use of continuity of f(x) and then he used differentiability of it and simply differentiated both sides of f(x).
@@priyansharma1512 That was what he showed in the video. But what he ASKED for in the thumbnail of the video - thus in the problem presentation - was ONLY differentiablity. ... Well: you could argue that there was the term "continuity" in the title. But it was not in the title as a part of the requirements, only as a description of the aspects which he will be talking about.
I really want to know how you want the problem be described. So it matches what I did in the video. We Never Stop Learning. I'll change the description if it makes sense.
@@PrimeNewtons Maybe in the thumbnail write : "f ' (x) differentiable" (the derivative has to be differentiable, not the original f)
...wait a moment: Could it be that the apostroph is already there but just not visible because the thumbnail being too small? ...No: Two videos earlier the apostroph is clearly visible... I really think it's missing in this thumbnail.
Oh, I think you misunderstand the problem. Ax+B is not the derivative of x²+x.