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Math Help - equation of a tangent

  1. #1
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    Exclamation equation of a tangent

    This is probably pretty easy but i am confused with how to continue

    The problem: Find an equation of the straight line having a slope \frac{1}{4} that is tangent to the curve y=\sqrt{x}

    I pluged some values in to a formula my prof. gave me \frac{f(x_{1}+h)-f(x_{1})}{h}

    Here is what I have so far:
    \frac{1}{4}=\lim_{h\rightarrow0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h}
    But i don't know where to go from here
    Thanks for looking
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  2. #2
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    Quote Originally Posted by nightrider456 View Post
    This is probably pretty easy but i am confused with how to continue

    The problem: Find an equation of the straight line having a slope \frac{1}{4} that is tangent to the curve y=\sqrt{x}

    I pluged some values in to a formula my prof. gave me \frac{f(x_{1}+h)-f(x_{1})}{h}

    Here is what I have so far:
    \frac{1}{4}=\lim_{h\rightarrow0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h}
    But i don't know where to go from here
    Thanks for looking
    \displaystyle \frac{1}{4} = \lim_{h\to 0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h} \cdot \frac{\sqrt{x_{1}+h}+\sqrt{x_{1}}}{\sqrt{x_{1}+h}+  \sqrt{x_{1}}}

    ring any bells?
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  3. #3
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    So I would then simplify from there?
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  4. #4
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    Quote Originally Posted by nightrider456 View Post
    So I would then simplify from there?
    I gave you the first step in finding the limit on the right hand side of the equation ... finish it by finding that limit, set it equal to 1/4 and solve for x_1.
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  5. #5
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    Are you required to use the definition of limit? If you know the "power rule", that the derivative of x^r, where r is a constant, is rx^{r-1}, then you can use that with r= \frac{1}{2}.
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