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Math Help - Differentiating from first principles

  1. #1
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    Differentiating from first principles

    Hey, could someone please help me with the second part of this question. Thanks!

    g(x) = ln x ; x > 1
    1.7x - C ; x != 1

    (a) Determine the value(s) of C for which this function is continuous at x = 1.
    (b) With each value of C found in (a), determine from first principles if g is differentiable at x = 1.

    For the first part I got C = 1.7, I'm unsure how to do part b.
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  2. #2
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    Re: Differentiating from first principles

    Learning some simple LaTeX to make the question readable would be nice... Is this your function?

    \displaystyle \begin{align*} g(x) = \begin{cases} \ln{(x)} \textrm{ if } x > 1 \\ 1.7^x - C \textrm{ if } x \neq 1 \end{cases} \end{align*}

    This is how it reads, though I don't see how it possibly could be, considering that would mean that for all \displaystyle \begin{align*} x > 1 \end{align*} you have the function equalling two different values...
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    Re: Differentiating from first principles

    Firstly - MATH1051 much? Leaving it a bit late to be asking questions on a forum don't you think?

    Quote Originally Posted by Prove It View Post
    Learning some simple LaTeX to make the question readable would be nice... Is this your function?

    \displaystyle \begin{align*} g(x) = \begin{cases} \ln{(x)} \textrm{ if } x > 1 \\ 1.7^x - C \textrm{ if } x \neq 1 \end{cases} \end{align*}

    This is how it reads, though I don't see how it possibly could be, considering that would mean that for all \displaystyle \begin{align*} x > 1 \end{align*} you have the function equalling two different values...
    Yes that is how it reads, I believe he meant to say =<, but for some reason wrote !=...I don't know either. So what I think he actually tried to ask reads like this;

    \displaystyle \begin{align*} g(x) = \begin{cases} \ln{(x)} \textrm{ if } x > 1 \\ 1.7^x - C \textrm{ if } x \leq 1 \end{cases} \end{align*}

    Okay so this is what I did, although I'm not sure it's the answer one would expect, so I would appreciate someone checking it so I don't mislead anyone;

    Alright so I assume you know the following formula:
     \frac{dy}{dx} = \lim_{h \rightarrow 0}  \frac{f(x+h) - f(x)}{h}

    Assuming yes then you just go from there, so;
     g(x+h) = 1.7^{x+h} - C \\ \frac{dy}{dx} = \lim_{h \rightarrow 0} \frac{1.7^{x+h} - C - (1.7^x - C)}{h} \\ \frac{dy}{dx} = \lim_{h \rightarrow 0} \frac{1.7^{x+h} - C - 1.7^x + C}{h}

    However normally when you use this formula there is a h on the nominator to cancel out the h on the denominator so you don't divide by 0. No such luck in this formula. I am quite certain this is the right way to go about solving it though and so, if you have any precognition you will see this formula is very quickly going to result in  \frac{0}{0}
    Last edited by Mrsupdup; May 1st 2013 at 08:19 PM.
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