- anonymous

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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

With what?

- anonymous

|dw:1326694446525:dw|

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.