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 our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

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

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

the following function is differentiable for all values of x.

|dw:1350070251929:dw|

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

a=1 ?

Oops. I figured out b = 1.

I can`t solve for a.

Because one this it must need to be is continuous

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

Should I derive both functions and solve for a?

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

Right that is one condition we must need though

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

So I go:
|dw:1350070954216:dw|

right? O_o .

No. Thats not right.

Never mind.

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

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

Right...

I tried that.

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

Let me try again. Maybe I made a mistake.

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

Is that an a on top ?

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

no it should say a on top

that is your issue

(ax+1)'=a

Ohh. Why?

Ohh....

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

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

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

Yep and the other side you have?

|dw:1350071580411:dw|

right?

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

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

I got a = pi/2

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

That is what I got :)

Gj again.