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Math Help - Find using the definition........the gradient function f '(x)

  1. #1
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    Post Find using the definition........the gradient function f '(x)

    Find using the defintion f '(c) lim = f(x) - f(c)
    x -> c _______
    x - c

    the gradient function f ' (x) for:

    a) f(x) = 1/x

    b)f(x) =
    √ x

    [My tutor told me to rationalize before doing the question for part b but how do i rationalize this]
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by sweetG View Post
    [My tutor told me to rationalize before doing the question for part b but how do i rationalize this]
    To rationalize the numerator of \frac{\sqrt{x}-\sqrt{c}}{x-c}, multiply and divide the fraction by the conjugate of \sqrt{x}-\sqrt{c} which is \sqrt{x}+\sqrt{c} :

    \frac{\sqrt{x}-\sqrt{c}}{x-c}=\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{(x-c)(\sqrt{x}+\sqrt{c})}=\ldots

    (notice that the numerator looks like (a-b)(a+b) which equals a^2-b^2)


    Edit : for the first question : \frac{f(x)-f(c)}{x-c}=\frac{\frac{1}{x}-\frac{1}{c}}{x-c}=\frac{\frac{c-x}{xc}}{x-c}=\frac{c-x}{xc(x-c)}=-\frac{1}{xc} hence f'(c) = ?
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  3. #3
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    Post

    Sorry I didn't realize that the definition for f '(c) was all over the place....sorry about that.


    For Part b is the answer:

    (x-c)/(x
    √x + x√c - c√x - c√c

    and for Part (a) how am I supposed to find f '(x)
    Sorry this question is really confusing me
    coz there are so many x's and c's.......
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by sweetG View Post
    [SIZE=1][SIZE=2]
    For Part b is the answer:

    (x-c)/(x√x + x√c - c√x - c√c)
    That's it but if you try computing f'(c)=\lim_{x\to c}\frac{x-c}{x\sqrt{x}+x\sqrt{c}-c\sqrt{x}-c\sqrt{c}} you'll see that this limit is an indeterminate form. (both the numerator and the denominator tend to 0) If you hadn't expanded the numerator, it would have given you

    \frac{\sqrt{x}-\sqrt{c}}{x-c}=\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{(x-c)(\sqrt{x}+\sqrt{c})}=\frac{x-c}{(x-c)\left(\sqrt{x}+\sqrt{c}\right)}=\frac{1}{\sqrt{x  }+\sqrt{c}}

    And f'(c)=\lim_{x\to c}\frac{1}{\sqrt{x}+\sqrt{c}} isn't an indeterminate form any longer. Can you find the value of this limit ?

    and for Part (a) how am I supposed to find f '(x)
    Sorry this question is really confusing me
    coz there are so many x's and c's.......
    The gradient of f taken at c is f'(c)=\lim_{x\to c} -\frac{1}{xc}= -\frac{1}{c\cdot c}=-\frac{1}{c^2} as long as c\neq 0. Thus \boxed{f'(x)=-\frac{1}{x^2}} for x\neq 0.
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  5. #5
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    Does f '(c) = 1/2√c

    which therefore makes f ' (x) = 1/2√x

    Is this right or have I made a mistake ?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by sweetG View Post
    Does f '(c) = 1/2√c

    which therefore makes f ' (x) = 1/2√x
    That's it !
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