Find a and b such that

- anonymous

Find a and b such that

- katieb

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

the following function is differentiable for all values of x.

- anonymous

|dw:1350070251929:dw|

- anonymous

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

Looking for something else?

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

## More answers

- myininaya

a=1 ?

- anonymous

Oops. I figured out b = 1.

- anonymous

I can`t solve for a.

- 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

Because one this it must need to be is continuous

- anonymous

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

- anonymous

Should I derive both functions and solve for a?

- anonymous

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

- myininaya

Right that is one condition we must need though

- myininaya

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

- anonymous

So I go:
|dw:1350070954216:dw|

- anonymous

right? O_o .

- anonymous

No. Thats not right.

- anonymous

Never mind.

- 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

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

- myininaya

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

- anonymous

Right...

- anonymous

I tried that.

- myininaya

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

- anonymous

Let me try again. Maybe I made a mistake.

- anonymous

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

- myininaya

Is that an a on top ?

- anonymous

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

- myininaya

no it should say a on top

- myininaya

that is your issue

- myininaya

(ax+1)'=a

- anonymous

Ohh. Why?

- anonymous

Ohh....

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

Yep and the other side you have?

- anonymous

|dw:1350071580411:dw|

- anonymous

right?

- anonymous

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

- myininaya

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

- anonymous

I got a = pi/2

- anonymous

- anonymous

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

- myininaya

That is what I got :)

- myininaya

Gj again.

Looking for something else?

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