A Nostalgic Problem Revisited With Sound 😁

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[@leaDR356]

I chose 1 by observation

[@doctorb9264]

Really like the emphasis on the graphs.

[@SyberMath]

Glad to hear that!

[@phill3986]

😊😊😊👍👍👍

[@farhansadik5423]

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

[@SyberMath]

Thanks! I use Notability on iPad

[@farhansadik5423]

@@SyberMath ah :) i'm too poor for that right now but i'll keep it in mind :))))

[@SidneiMV]

WOW! Domain! Domain!!

[@SidneiMV]

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

[@hazem1el_abed]

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

[@BlaqRaq]

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

[@YouTube_username_not_found]

The first method is definitely more elegant!

[@SyberMath]

Glad you think so!

[@johns.8246]

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

[@DonEnsley-mathdrum]

X=1. The negative solution of -1 doesn't work.

[@fahrenheit2101]

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.

[@rakenzarnsworld2]

x = 1

[@tunistick8044]

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

[@SyberMath]

ooopsies

[@moeberry8226]

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.

[@YouTube_username_not_found]

Sooo, does the inverse function have a critical point at 0, or this point doesn't count because it's not in the domain?

[@moeberry8226]

@@KZbin_username_not_found what are you talking about which function? Log_5(x) has an inverse but 5^(sqrt(1-x^2)) does not.

[@YouTube_username_not_found]

@@moeberry8226 Excuse me. I meant the reciprocal function.

[@YouTube_username_not_found]

The function f such that f(x) = 1/x

[@moeberry8226]

@@KZbin_username_not_found that has nothing to do with my comment or any function related to the video.

[@peskarr]

too easy, solved with halfcircle in 1 minute