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help!!!!!!!!!!!!!!

Mathematics
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With what?
|dw:1326694446525:dw|
if f(x)is continuous at x=0...............find a,b

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Other answers:

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, \[\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})\]
so this means also we need \[\lim_{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim_{x \rightarrow \frac{\pi}{2}^-}f(x)\]
so for x=pi/2 f(x)=a we will use this later!
myini: the question says that f(x) us continuous at x=0
\[f(\frac{\pi}{2})=a\]
oh that's interesting we wouldn't care about a then
or b
because that part says x>pi/2
0 is less than pi/2
k
i think the question had a type-o
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, \[\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}=\lim \frac{-3\sin^2x\cos x}{-6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a\]
because there is nothing to do if they aren't talking about pi/2
Similarly, solve for b.
and see imperialist ignored that continuous part at 0 lol
also*
but yeah i think they meant at pi/2
for b ?
Yeah, me too, I didn't even notice the zero part.
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.
i was trying to make the function continuous everywhere
@imper:but the question says that the function is continuous at x=0
he knows
It's a typo, it's impossible to make it continuous at x=0.
You already know how x is defined at zero, that can't be changed, no matter what you make a or b.
its already continuous at 0
cos(0)=1
True dat, my bad, I didn't look back at the question and assumed sin was on the bottom.
thats why the question makes no sense the work is already done i believe it meant what me and imperialist were trying to do
k
thanks guys !!!!!!!!!!!!!!!!
np :)
Glad to help :)
;)

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