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

- anonymous

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

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- anonymous

With what?

- anonymous

|dw:1326694446525:dw|

- anonymous

if f(x)is continuous at x=0...............find a,b

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## More answers

- myininaya

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

so this means also we need
\[\lim_{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim_{x \rightarrow \frac{\pi}{2}^-}f(x)\]

- myininaya

so for x=pi/2 f(x)=a
we will use this later!

- anonymous

myini: the question says that f(x) us continuous at x=0

- myininaya

\[f(\frac{\pi}{2})=a\]

- myininaya

oh that's interesting we wouldn't care about a then

- myininaya

or b

- myininaya

because that part says x>pi/2

- myininaya

0 is less than pi/2

- anonymous

k

- myininaya

i think the question had a type-o

- anonymous

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

because there is nothing to do if they aren't talking about pi/2

- anonymous

Similarly, solve for b.

- myininaya

and see imperialist ignored that continuous part at 0 lol

- myininaya

also*

- myininaya

but yeah i think they meant at pi/2

- anonymous

for b ?

- anonymous

Yeah, me too, I didn't even notice the zero part.

- anonymous

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

i was trying to make the function continuous everywhere

- anonymous

@imper:but the question says that the function is continuous at x=0

- myininaya

he knows

- anonymous

It's a typo, it's impossible to make it continuous at x=0.

- anonymous

You already know how x is defined at zero, that can't be changed, no matter what you make a or b.

- myininaya

its already continuous at 0

- myininaya

cos(0)=1

- anonymous

True dat, my bad, I didn't look back at the question and assumed sin was on the bottom.

- myininaya

thats why the question makes no sense the work is already done
i believe it meant what me and imperialist were trying to do

- anonymous

k

- anonymous

thanks guys !!!!!!!!!!!!!!!!

- myininaya

np :)

- anonymous

Glad to help :)

- anonymous

;)

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