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Math Help - domain of derivative

  1. #1
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    domain of derivative

    Domain of f(x)=sqrt(1+2x)
    1+2x>=0
    x>=-1/2
    domain:: (-1/2,+inf)


    and
    domain of f '(x)=1/sqrt(1+2x).........do not understand to do this one..
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  2. #2
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    Hello, Gracy!

    A minor (but important) change in your first answer . . .


    Domain of f(x) = \sqrt{1+2x}

    1 + 2x \:\geq \:0\quad\Rightarrow\quad x\:\geq \:-\frac{1}{2}

    Domain: . \left[-\frac{1}{2},\:\infty\right)
    . . . . . . .

    We include the -\frac{1}{2}



    Domain of f'(x) \:=\:\frac{1}{\sqrt{1+2x}}

    That prime-mark is distracting, but we can ignore it.

    Since 1 + 2x is under a square root, it must be positive or 0.
    But since it is in the denominator, it cannot equal 0.

    Hence, we have: . 1 + 2x\:>\:0\quad\Rightarrow\quad x \:>\:-\frac{1}{2}

    Domain: . \left(-\frac{1}{2},\:\infty\right)

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  3. #3
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    Basically whenever you have a continous function on [a,b] and differenciable at (a,b) then the derivative does not exist at 'a' and 'b'. Do you see why? One of the conditions that implies a function is not differenciable is when it fails to be countinous. Note whenever you have a continous function on [a,b] then it is continous from the right at 'a' and continous from the left at 'b' but it is not fully continous.
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