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.

Need help on differentiables. Determining the values of constants

Mathematics
See more answers at brainly.com
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

\[f(x)=\left[\begin{matrix}6x^3 & x \le 1 \\ Bx+C & 1 < x\end{matrix}\right]\]
How do I find B and C
1\(^\pm\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Take the derivative of the function (each function separately), then make the derivative continuous.
Ok so i just take the derivitive of both: \[f'(x)=\left[\begin{matrix}18x^2 & x<=1 \\ B & 1
Now insert \(x=1\) for both parts, and solve for \(B\)
ok so: \[f'(x)=18(1)^2=18 = B\] Correct?
Yes.
Now you need to make the original function continuous.
So: \[f(x)= 18(1)+C<1\]
So c=-17?
No, the other equation to equal it to is \(6(1)^3\)
Oh ok so then f(x)=6 so then: \[18(1)+C=6\] Then I do this \[C=-12\]
Correct?
C and B values are both correct. Looking over your work though, I'm not sure you understand what wio was trying to get you to do.
I'm saying your work could use a little work. :p But yes the answers are right.
Thats fine as long as your trying to help out :)
So when I see a problem like this I take the derivative of both equations and make the darivative continuous. Then I go back and plug in B and make just f(x) continous.
Basically they were trying to make the function f continue and also trying to make the function f' continuous Make the function f' continuous will make the function f differential if we choose the right values for B and C of course. So when we looked at f We did the left and right limit as x approaches 1. Set both the left limit expression and the right limit expression equal. We notice here we don't have enough info to solve for B and C, but we do have an equation relating C and B from the left limit expression=right limit expression equation. f(x)=6x^3 f(x)=Bx+C _________________|_________________ 1 You want both sides of 1 to be equal as they approach 1. So left limit expression is 6(1)^3 and the right limit expression is B(1)+C Remember you need to set these equal giving you that relationship between B and C that was talking about. Now we also need to find f' f'(x)=18x^2 f'(x)=B ----------------|---------------- 1 You do the same here that you did for the original function f. You set the left limit expression equal to the right limit expression as x approaches 1.
I'm just making sure you understand so you can tackle problems like this in the future.
Ya it makes much more sense now. Thank you myininaya!

Not the answer you are looking for?

Search for more explanations.

Ask your own question