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

How do I find B and C

1\(^\pm\)

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?

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 :)

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!