anonymous
  • anonymous
Find a and b such that
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
the following function is differentiable for all values of x.
anonymous
  • anonymous
|dw:1350070251929:dw|
anonymous
  • anonymous
I figured out a= 1 but I can't figure out what b is equal to.

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myininaya
  • myininaya
a=1 ?
anonymous
  • anonymous
Oops. I figured out b = 1.
anonymous
  • anonymous
I can`t solve for a.
myininaya
  • 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.
myininaya
  • myininaya
Because one this it must need to be is continuous
anonymous
  • anonymous
Yep . I set the both as lim x-> 0 and I got that b = 1. I can't solve for a however.
anonymous
  • anonymous
Should I derive both functions and solve for a?
anonymous
  • anonymous
But that just means it's continuous. Not differentiable.
myininaya
  • myininaya
Right that is one condition we must need though
myininaya
  • myininaya
We need it to run smoothly when they meet (no sharp corners)... hmm... thinking....
anonymous
  • anonymous
So I go: |dw:1350070954216:dw|
anonymous
  • anonymous
right? O_o .
anonymous
  • anonymous
No. Thats not right.
anonymous
  • anonymous
Never mind.
myininaya
  • 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
anonymous
  • anonymous
Yep. I knew that. But unless I know what a is equal to I can't solve for the derivative at x=0.
myininaya
  • myininaya
\[(\arctan(ax+1))'|_{x=0}=(\frac{\pi}{4} e^{\sin(x)})'|_{x=0}\]
anonymous
  • anonymous
Right...
anonymous
  • anonymous
I tried that.
myininaya
  • myininaya
Find derivative of both sides Set x=0 And solve you are done
anonymous
  • anonymous
Let me try again. Maybe I made a mistake.
anonymous
  • anonymous
The derivative of arctan(ax+1) |dw:1350071299130:dw| right?
myininaya
  • myininaya
Is that an a on top ?
anonymous
  • anonymous
It shoud say 1/(1+(ax+1)^2)
myininaya
  • myininaya
no it should say a on top
myininaya
  • myininaya
that is your issue
myininaya
  • myininaya
(ax+1)'=a
anonymous
  • anonymous
Ohh. Why?
anonymous
  • anonymous
Ohh....
anonymous
  • anonymous
haha... Chain rule.... oops.
myininaya
  • myininaya
\[(\arctan(u(x))'=\frac{u'(x)}{{1+(u(x))^2}}\]
anonymous
  • anonymous
so it would be \[\frac{ a }{ 1+(ax+1)^2 }\]
myininaya
  • myininaya
Yep and the other side you have?
anonymous
  • anonymous
|dw:1350071580411:dw|
anonymous
  • anonymous
right?
anonymous
  • anonymous
I really appreciate the help by the way. Thanks a lot.
myininaya
  • myininaya
I think you got it from here :) Great job.
anonymous
  • anonymous
I got a = pi/2
anonymous
  • anonymous
anonymous
  • anonymous
Thanks for making this so easy to understand! :) .
myininaya
  • myininaya
That is what I got :)
myininaya
  • myininaya
Gj again.

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