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Math Help - derivative

  1. #1
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    derivative

    suppose f(x)= sqrt of 1+3x, x>or equal to 5
    k(x^2-5), x<5
    for what value of k is continuous? For what value of k is f differentiable?

    not even sure where to start i think i am on dealing with the second equation in this piecewise function

    i know the answers are f is continuous if k=1/5 and f is differentiable if k=3/80
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  2. #2
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    f\left( x \right) = \left\{ \begin{gathered}<br />
  \sqrt {1 + 3x} ,{\text{       if  }}x \geqslant 5 \hfill \\<br />
  k\left( {x^2  - 5} \right){\text{,    if  }}x < 5 \hfill \\ <br />
\end{gathered}  \right. \hfill \\

     {\text{for }}f\left( x \right){\text{ to be continuous, it should be continuous at }}x = 5. \hfill \\


    {\text{For }}f\left( x \right){\text{ to be continuous at }}x = 5, \hfill \\

     \mathop {\lim }\limits_{x \to 5 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 + } f\left( x \right) = f\left( 5 \right) \hfill \\


    {\text{Now,}} \hfill \\

    \mathop {\lim }\limits_{x \to 5 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 - } k\left( {x^2  - 5} \right) = k\left( {5^2  - 5} \right) = 20k \hfill \\

    \mathop {\lim }\limits_{x \to 5 + } f\left( x \right) = \mathop {\lim }\limits_{x \to 5 + } \sqrt {1 + 3x}  = \sqrt {1 + 3\left( 5 \right)}  = 4 \hfill \\


    \Rightarrow 20k = 4 \hfill \\

    \Rightarrow k = \frac{1}<br />
{5} \hfill \\

    f(x) is continuous when k = 1/5
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  3. #3
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    thanks how about finding the differentiable being 3/80
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  4. #4
    Member Jason Bourne's Avatar
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    Quote Originally Posted by vinson24 View Post
    thanks how about finding the differentiable being 3/80
    f'\left( x \right) = \left\{ \begin{gathered}<br />
  \frac{3}{2\sqrt {1 + 3x}}  \hfill \\<br />
  2kx \hfill \\ <br />
\end{gathered}  \right. \hfill \\

    at x=5 , I think this is similar to before but using the above, you should look over any notes you have on continuity and differentiability.
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