Need help on differentiables. Determining the values of constants

- anonymous

Need help on differentiables. Determining the values of constants

- katieb

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- anonymous

\[f(x)=\left[\begin{matrix}6x^3 & x \le 1 \\ Bx+C & 1 < x\end{matrix}\right]\]

- anonymous

How do I find B and C

- abb0t

1\(^\pm\)

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## More answers

- anonymous

Take the derivative of the function (each function separately), then make the derivative continuous.

- anonymous

Ok so i just take the derivitive of both:
\[f'(x)=\left[\begin{matrix}18x^2 & x<=1 \\ B & 1

- anonymous

Now insert \(x=1\) for both parts, and solve for \(B\)

- anonymous

ok so:
\[f'(x)=18(1)^2=18 = B\]
Correct?

- anonymous

Yes.

- anonymous

Now you need to make the original function continuous.

- anonymous

So:
\[f(x)= 18(1)+C<1\]

- anonymous

So c=-17?

- anonymous

No, the other equation to equal it to is \(6(1)^3\)

- anonymous

Oh ok so then f(x)=6 so then:
\[18(1)+C=6\]
Then I do this
\[C=-12\]

- anonymous

Correct?

- myininaya

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.

- myininaya

I'm saying your work could use a little work. :p
But yes the answers are right.

- anonymous

Thats fine as long as your trying to help out :)

- anonymous

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.

- myininaya

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.

- myininaya

I'm just making sure you understand so you can tackle problems like this in the future.

- anonymous

Ya it makes much more sense now. Thank you myininaya!

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