DHASHNI
help!!!!!!!!!!!!!!
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pratu043
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With what?
DHASHNI
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|dw:1326694446525:dw|
DHASHNI
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if f(x)is continuous at x=0...............find a,b
myininaya
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Ok so this means we want the function to have the following properties:
\[f(\frac{\pi}{2}) \text{ exist}\]
\[\lim_{x \rightarrow \frac{\pi}{2}}f(x) \text{ exist}\]
Finally,
\[\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})\]
myininaya
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so this means also we need
\[\lim_{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim_{x \rightarrow \frac{\pi}{2}^-}f(x)\]
myininaya
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so for x=pi/2 f(x)=a
we will use this later!
DHASHNI
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myini: the question says that f(x) us continuous at x=0
myininaya
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\[f(\frac{\pi}{2})=a\]
myininaya
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oh that's interesting we wouldn't care about a then
myininaya
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or b
myininaya
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because that part says x>pi/2
myininaya
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0 is less than pi/2
DHASHNI
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k
myininaya
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i think the question had a type-o
imperialist
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Just take a couple of limits using l'hopitals rule. For the first one since \[1-\sin^3(\pi/2)=3\cos^2(\pi/2)=0\] the constraints of l'hopital's rule applies for the left limit. Thus, \[\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}=\lim \frac{-3\sin^2x\cos x}{-6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a\]
myininaya
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because there is nothing to do if they aren't talking about pi/2
imperialist
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Similarly, solve for b.
myininaya
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and see imperialist ignored that continuous part at 0 lol
myininaya
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also*
myininaya
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but yeah i think they meant at pi/2
DHASHNI
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for b ?
imperialist
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Yeah, me too, I didn't even notice the zero part.
imperialist
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For b, you just do the same thing, except you take the limit as x approaches pi/2 from the right regarding the second half of the piecewise function. Set that equal to 1/2, since you know the limit must be that, and you are done.
myininaya
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i was trying to make the function continuous everywhere
DHASHNI
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@imper:but the question says that the function is continuous at x=0
myininaya
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he knows
imperialist
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It's a typo, it's impossible to make it continuous at x=0.
imperialist
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You already know how x is defined at zero, that can't be changed, no matter what you make a or b.
myininaya
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its already continuous at 0
myininaya
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cos(0)=1
imperialist
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True dat, my bad, I didn't look back at the question and assumed sin was on the bottom.
myininaya
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thats why the question makes no sense the work is already done
i believe it meant what me and imperialist were trying to do
DHASHNI
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k
DHASHNI
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thanks guys !!!!!!!!!!!!!!!!
myininaya
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np :)
imperialist
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Glad to help :)
DHASHNI
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;)