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Matthew071 Group Title

Need help on differentiables. Determining the values of constants

  • 11 months ago
  • 11 months ago

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  1. Matthew071 Group Title
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    \[f(x)=\left[\begin{matrix}6x^3 & x \le 1 \\ Bx+C & 1 < x\end{matrix}\right]\]

    • 11 months ago
  2. Matthew071 Group Title
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    How do I find B and C

    • 11 months ago
  3. abb0t Group Title
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    1\(^\pm\)

    • 11 months ago
  4. wio Group Title
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    Take the derivative of the function (each function separately), then make the derivative continuous.

    • 11 months ago
  5. Matthew071 Group Title
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    Ok so i just take the derivitive of both: \[f'(x)=\left[\begin{matrix}18x^2 & x<=1 \\ B & 1<x\end{matrix}\right]\]

    • 11 months ago
  6. wio Group Title
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    Now insert \(x=1\) for both parts, and solve for \(B\)

    • 11 months ago
  7. Matthew071 Group Title
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    ok so: \[f'(x)=18(1)^2=18 = B\] Correct?

    • 11 months ago
  8. wio Group Title
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    Yes.

    • 11 months ago
  9. wio Group Title
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    Now you need to make the original function continuous.

    • 11 months ago
  10. Matthew071 Group Title
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    So: \[f(x)= 18(1)+C<1\]

    • 11 months ago
  11. Matthew071 Group Title
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    So c=-17?

    • 11 months ago
  12. wio Group Title
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    No, the other equation to equal it to is \(6(1)^3\)

    • 11 months ago
  13. Matthew071 Group Title
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    Oh ok so then f(x)=6 so then: \[18(1)+C=6\] Then I do this \[C=-12\]

    • 11 months ago
  14. Matthew071 Group Title
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    Correct?

    • 11 months ago
  15. myininaya Group Title
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    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.

    • 11 months ago
  16. myininaya Group Title
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    I'm saying your work could use a little work. :p But yes the answers are right.

    • 11 months ago
  17. Matthew071 Group Title
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    Thats fine as long as your trying to help out :)

    • 11 months ago
  18. Matthew071 Group Title
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    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.

    • 11 months ago
  19. myininaya Group Title
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    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.

    • 11 months ago
  20. myininaya Group Title
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    I'm just making sure you understand so you can tackle problems like this in the future.

    • 11 months ago
  21. Matthew071 Group Title
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    Ya it makes much more sense now. Thank you myininaya!

    • 11 months ago
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