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Thread: Domain & first derivative

  1. #1
    Junior Member
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    Domain & first derivative

    $\displaystyle
    \begin{array}{l}
    f(x) = \sqrt {1 - \sqrt {2 - \sqrt {3 - x} } } \\
    \\
    find \\
    1)\;domain\;of\;f(x) \\
    2)\;f'(x) \\
    \end{array}


    $
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  2. #2
    MHF Contributor

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    First, of course, 3- x must be non-negative in order for that square root to be a real number: $\displaystyle 3- x\ge 0$ so $\displaystyle x\le 3$. And once we have that we must have $\displaystyle 2-\sqrt{3- x}\ge 0$. That is the same as saying that $\displaystyle 2\ge \sqrt{3- x}$ or $\displaystyle 4\ge 3- x$, $\displaystyle x\ge -1$. So far then, $\displaystyle -1\le x\le 3$. Now, we need to have $\displaystyle 1- \sqrt{2- \sqrt{3- x}}\ge 0$. Can you finish that?

    As for the derivative, this function is $\displaystyle y= (1- (2- (3- x)^{1/2})^{1/2})^{1/2}$. Use the "power rule" and the "chain rule".
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  3. #3
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    $\displaystyle f^{'}(x)={\frac{-1}{8\sqrt{1-\sqrt{2-\sqrt{3-x}}}\sqrt{2-\sqrt{3-x}}\sqrt{3-x}}$,
    and then you can simplify the answer
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