A Nostalgic Problem Revisited With Sound 😁 

I chose 1 by observation

Really like the emphasis on the graphs.

8:28 intervals should be [-1,0] and [0,1]

I think, you should put y² instead of 1-x² y² + 5²∧y = 1 lim ₓ→₀( f ) and lim ₓ→-∞( f ) first one is ok then y² =0 so that x =1 also

It (exponentialize) will, more likely, show up in the dictionary on Friday. Remember, the different departments it has to pass through?

epic video! Would you tell me what application you use for making these? I recently got a drawing tablet and want to imitate you in a mirror lol

x = cosu => x² = cos²u 1 - x² = sin²u => √(1 - x²) = sinu 5^sinu = cosu => 0 < cosu 0 1

The first method is definitely more elegant!

1 is an obvious solution, and a graph sketch confirms it's the only one, since sqrt(1 - x^2) would have a upper semicircular graph, and exponentiating preserves order, so the graph shape would still be "up then down", enough to ensure 1 is the only solution. Though and algebraic approach escapes me, so ... time to watch.

If p is prime, when is 7^(3p) - 2^(3p) a multiple of p?

WOW! Domain! Domain!!

😊😊😊👍👍👍

8:26 You have made a little mistake in the domain in the table of variations of f. It's ]-1;1[ not IR

too easy, solved with halfcircle in 1 minute

I want to point out that critical points are not just were the function has a zero slope but rather where the slope is also undefined and that can happen at points of discontinuity, a sharp cusp or a vertical tangent.

x = 1