anonymous 4 years ago Find a and b such that

1. anonymous

the following function is differentiable for all values of x.

2. anonymous

|dw:1350070251929:dw|

3. anonymous

I figured out a= 1 but I can't figure out what b is equal to.

4. myininaya

a=1 ?

5. anonymous

Oops. I figured out b = 1.

6. anonymous

I can`t solve for a.

7. myininaya

Yep. Now that is right so far. You just set the limx->0 for both of them and solve for a b. Good job.

8. myininaya

Because one this it must need to be is continuous

9. anonymous

Yep . I set the both as lim x-> 0 and I got that b = 1. I can't solve for a however.

10. anonymous

Should I derive both functions and solve for a?

11. anonymous

But that just means it's continuous. Not differentiable.

12. myininaya

Right that is one condition we must need though

13. myininaya

We need it to run smoothly when they meet (no sharp corners)... hmm... thinking....

14. anonymous

So I go: |dw:1350070954216:dw|

15. anonymous

right? O_o .

16. anonymous

No. Thats not right.

17. anonymous

Never mind.

18. myininaya

Ok...how about this : What do you think of both parts of the function having the same exact slope at x=0 This will give us that smooth transition at x=0 that i'm talking about

19. anonymous

Yep. I knew that. But unless I know what a is equal to I can't solve for the derivative at x=0.

20. myininaya

$(\arctan(ax+1))'|_{x=0}=(\frac{\pi}{4} e^{\sin(x)})'|_{x=0}$

21. anonymous

Right...

22. anonymous

I tried that.

23. myininaya

Find derivative of both sides Set x=0 And solve you are done

24. anonymous

Let me try again. Maybe I made a mistake.

25. anonymous

The derivative of arctan(ax+1) |dw:1350071299130:dw| right?

26. myininaya

Is that an a on top ?

27. anonymous

It shoud say 1/(1+(ax+1)^2)

28. myininaya

no it should say a on top

29. myininaya

30. myininaya

(ax+1)'=a

31. anonymous

Ohh. Why?

32. anonymous

Ohh....

33. anonymous

haha... Chain rule.... oops.

34. myininaya

$(\arctan(u(x))'=\frac{u'(x)}{{1+(u(x))^2}}$

35. anonymous

so it would be $\frac{ a }{ 1+(ax+1)^2 }$

36. myininaya

Yep and the other side you have?

37. anonymous

|dw:1350071580411:dw|

38. anonymous

right?

39. anonymous

I really appreciate the help by the way. Thanks a lot.

40. myininaya

I think you got it from here :) Great job.

41. anonymous

I got a = pi/2

42. anonymous

@myininaya .

43. anonymous

Thanks for making this so easy to understand! :) .

44. myininaya

That is what I got :)

45. myininaya

Gj again.