## DHASHNI Group Title help!!!!!!!!!!!!!! 2 years ago 2 years ago

1. pratu043 Group Title

With what?

2. DHASHNI Group Title

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3. DHASHNI Group Title

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

4. myininaya Group Title

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, $\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})$

5. myininaya Group Title

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

6. myininaya Group Title

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

7. DHASHNI Group Title

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

8. myininaya Group Title

$f(\frac{\pi}{2})=a$

9. myininaya Group Title

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

10. myininaya Group Title

or b

11. myininaya Group Title

because that part says x>pi/2

12. myininaya Group Title

0 is less than pi/2

13. DHASHNI Group Title

k

14. myininaya Group Title

i think the question had a type-o

15. imperialist Group Title

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, $\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}=\lim \frac{-3\sin^2x\cos x}{-6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a$

16. myininaya Group Title

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

17. imperialist Group Title

Similarly, solve for b.

18. myininaya Group Title

and see imperialist ignored that continuous part at 0 lol

19. myininaya Group Title

also*

20. myininaya Group Title

but yeah i think they meant at pi/2

21. DHASHNI Group Title

for b ?

22. imperialist Group Title

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

23. imperialist Group Title

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.

24. myininaya Group Title

i was trying to make the function continuous everywhere

25. DHASHNI Group Title

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

26. myininaya Group Title

he knows

27. imperialist Group Title

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

28. imperialist Group Title

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

29. myininaya Group Title

30. myininaya Group Title

cos(0)=1

31. imperialist Group Title

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

32. myininaya Group Title

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

33. DHASHNI Group Title

k

34. DHASHNI Group Title

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

35. myininaya Group Title

np :)

36. imperialist Group Title