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Need help on differentiables. Determining the values of constants
 6 months ago
 6 months ago
Need help on differentiables. Determining the values of constants
 6 months ago
 6 months ago

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Matthew071Best ResponseYou've already chosen the best response.0
\[f(x)=\left[\begin{matrix}6x^3 & x \le 1 \\ Bx+C & 1 < x\end{matrix}\right]\]
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
How do I find B and C
 6 months ago

wioBest ResponseYou've already chosen the best response.1
Take the derivative of the function (each function separately), then make the derivative continuous.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Ok so i just take the derivitive of both: \[f'(x)=\left[\begin{matrix}18x^2 & x<=1 \\ B & 1<x\end{matrix}\right]\]
 6 months ago

wioBest ResponseYou've already chosen the best response.1
Now insert \(x=1\) for both parts, and solve for \(B\)
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
ok so: \[f'(x)=18(1)^2=18 = B\] Correct?
 6 months ago

wioBest ResponseYou've already chosen the best response.1
Now you need to make the original function continuous.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
So: \[f(x)= 18(1)+C<1\]
 6 months ago

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

Matthew071Best ResponseYou've already chosen the best response.0
Oh ok so then f(x)=6 so then: \[18(1)+C=6\] Then I do this \[C=12\]
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
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.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
I'm saying your work could use a little work. :p But yes the answers are right.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Thats fine as long as your trying to help out :)
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
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.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
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.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
I'm just making sure you understand so you can tackle problems like this in the future.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Ya it makes much more sense now. Thank you myininaya!
 6 months ago
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