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The x=1 solution can be seen by inspection.
If 0
Did I write it the other way around? 😮
@@SyberMath Yes, you wrote LHS < RHS for the 0-1 interval. I spotted the same mistake.
For x < 0, the LHS is greater than 2 + 3 since the exponents are positive, and the RHS is less than 1 since the exponent (x^5) is negative
That only holds for x=
I'd like a mathematical deduction, solving the equation...
at 7:25 you said that the term 2^(x^2)/2^(x^5) is bounded by 2/5 which is true but seems to come out of nowhere (the term is bounded by 1 even). My guess is that for the denominator you calculated with 5^(x^5) instead and got the 2/5 from there
Yes, it appears he meant to write 2^(1-x^2) < 5^(1-x^5), from which we find(2^(x^2))/(5^(x^5)) > 2/5
From Romania with love. Maybe SyberMath speech is from there?
😍❤️ with love from The United States...
x = 1
The x=1 solution can be seen by inspection.
If 0
Did I write it the other way around? 😮
@@SyberMath Yes, you wrote LHS < RHS for the 0-1 interval. I spotted the same mistake.
For x < 0, the LHS is greater than 2 + 3 since the exponents are positive, and the RHS is less than 1 since the exponent (x^5) is negative
That only holds for x=
I'd like a mathematical deduction, solving the equation...
at 7:25 you said that the term 2^(x^2)/2^(x^5) is bounded by 2/5 which is true but seems to come out of nowhere (the term is bounded by 1 even). My guess is that for the denominator you calculated with 5^(x^5) instead and got the 2/5 from there
Yes, it appears he meant to write 2^(1-x^2) < 5^(1-x^5), from which we find
(2^(x^2))/(5^(x^5)) > 2/5
From Romania with love. Maybe SyberMath speech is from there?
😍❤️ with love from The United States...
x = 1