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Thread: Difference Quotient

  1. #1
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    Difference Quotient

    I am new here but I have a question about this difference quotient. Hope you can help me!

    f(x)=√5x
    f(x)-f(5)/x-5

    didn't know how to show it but f(x)-f(5) are all over x-5
    So thank you if you can help me figure this out! I know the answer but I am not sure how to get to it....
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  2. #2
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    Quote Originally Posted by TimelordGurl View Post
    I am new here but I have a question about this difference quotient. Hope you can help me!

    f(x)=√5x
    f(x)-f(5)/x-5

    didn't know how to show it but f(x)-f(5) are all over x-5
    So thank you if you can help me figure this out! I know the answer but I am not sure how to get to it....
    Okay, so this is $\displaystyle \frac{f(x)- f(5)}{x-5}$. Now replace "f(x)" and "f(5)" by this specific function.

    But is $\displaystyle f(x)= (\sqrt{5})x$ or is $\displaystyle f(x)= \sqrt{5x}$?

    If the first, $\displaystyle \frac{f(x)- f(5)}{x-5}= \frac{(\sqrt{5}x- \sqrt{5}}{x-5}$$\displaystyle = \frac{\sqrt{5}(x-5)}{x-5}= \sqrt{5}$.
    (Which you should have expected- $\displaystyle y= (\sqrt{5})x$ is a linear function with slope $\displaystyle \sqrt{5}$.)

    If the second, $\displaystyle frac{f(x)- f(5)}{x-5}= \frac{\sqrt{5x}- \sqrt{5}}{x-5}$ which is harder to simplify. Try multiplying both numerator and denominator by $\displaystyle \sqrt{5x}+\sqrt{5}$.
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