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Matthew071
 one year ago
Need help on differentiables. Determining the values of constants
Matthew071
 one year ago
Need help on differentiables. Determining the values of constants

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Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=\left[\begin{matrix}6x^3 & x \le 1 \\ Bx+C & 1 < x\end{matrix}\right]\]

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0How do I find B and C

wio
 one year ago
Best ResponseYou've already chosen the best response.1Take the derivative of the function (each function separately), then make the derivative continuous.

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0Ok so i just take the derivitive of both: \[f'(x)=\left[\begin{matrix}18x^2 & x<=1 \\ B & 1<x\end{matrix}\right]\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Now insert \(x=1\) for both parts, and solve for \(B\)

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0ok so: \[f'(x)=18(1)^2=18 = B\] Correct?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Now you need to make the original function continuous.

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0So: \[f(x)= 18(1)+C<1\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1No, the other equation to equal it to is \(6(1)^3\)

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok so then f(x)=6 so then: \[18(1)+C=6\] Then I do this \[C=12\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1C 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I'm saying your work could use a little work. :p But yes the answers are right.

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0Thats fine as long as your trying to help out :)

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Basically 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I'm just making sure you understand so you can tackle problems like this in the future.

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0Ya it makes much more sense now. Thank you myininaya!
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