Differentiability and continuity of a piecewise function

In this video, I used the differentiability and continuity properties of a piece-wise function to determine the unkown constants.

[stigastondogg]

“Piecewise Smooth” was going to be my rap name, but that career never took off.

[johnnolen]

"A = -5. 'No way!' "

He's teaching the concepts, but he's still astonished by the answer. You gotta love the enthusiasm. 😇

[davidcolf]

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!

[KasahunTefera-xcni]

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.

[gurbani]

i have known how to solve these kind of problems but thanks to you i got in depth meaning and knowledge :)

[JourneyThroughMath]

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

[mikefochtman]

Great follow up to your other video about showing a continuous function.

[spoopy]

I love your energy so much, you seem to have so much passion for math ❤

[cameronconely]

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.

[samarthpriyadarshi]

I being in a low class am able to study these due to this channel❤

[marcgriselhubert]

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)

[kevinmadden]

Since differentiability implies continuity, it might be a good idea to work the derivative part before you work the limit part .

[jacobgoldman]

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.

[WhiteGandalfs]

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.