asianese
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Good .If a function is odd, then f(x)=-f(-x). Differentiating both sides, we get f'(x) = f'(-x) therefore the derivative is even.
If a function is even, then f(x)=f(-x). Differentiating both sides, we get f'(x) = -f'(-x) therefore the derivative is odd.
Yes. f(x)=k, k!=0 is one example. f(x)=x+1 is another.
Find the error (if any) in the following working:
Indeed any odd function plus a (non zero) constant is an example of a non-odd function with an even derivative.If a function is odd, then f(x)=-f(-x). Differentiating both sides, we get f'(x) = f'(-x) therefore the derivative is even.
If a function is even, then f(x)=f(-x). Differentiating both sides, we get f'(x) = -f'(-x) therefore the derivative is odd.
Yes. f(x)=k, k!=0 is one example. f(x)=x+1 is another.
Essentially. The main point is that in the changes of variables theorem you need a 1-1 correspondence between your old variable and your new variable...a point often glossed over by teachers.
EDIT: Done. Is this correct?
Can you explain this what you mean by 1-1 correspondence?Essentially. The main point is that in the changes of variables theorem you need a 1-1 correspondence between your old variable and your new variable...a point often glossed over by teachers.
Ok yep I get it.A bijection (wikipedia the term).
The problem here is that as x ranges from -1 to 1, u hits every value in (0,1] TWICE instead of once. You resolved this by splitting the integral into two intervals on which u doesn't have this behaviour.
(Also, naively looking at what u is at the endpoints of the integration in x and making the substitution, using these u-values as the new endpoints is flawed for this reason. The mapping from x->u takes the interval [-1,1] to the interval [0,1], not to the single number 1.)
The amount of words starting with A, B, C, D is:We wish to construct three-letter words from 26 letter of the english alphabet. How many possibilities are there if:
The word must come before EGG in alphabetical order?
You are missing EEConsider 3-letter string starting with A, B, C, D. There are 4*26^2=2704. There are 26 for each EA, EB, EC, ED, EF, so 2704+26*5=2834. Then there's EGA,EGB,...,EGG, 7 of these. So 2841 in total.
(edit: assuming no counting mistake)
Its the DThe amount of words starting with A, B, C, D is:
The amount of words starting with E, middle of A, B, C, D, E or F is
The amount of words starting with E, middle of G before EGG is 6.
So final answer:
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Since sin^2(0) = cos^2 (90), sin^2 (3) = cos^2 (87), sin^2 (6) = cos^2 (84), sin^2 (93) = cos^2 (177), ...