Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find a and b such that

Mathematics
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.

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

a=1 ?
Oops. I figured out b = 1.
I can`t solve for a.
Yep. Now that is right so far. You just set the limx->0 for both of them and solve for a b. Good job.
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.
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
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question