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I was wondering how you would approach this question: Estimate $e^x$ at $x = ln(10) + 0.1$, using the method of small increments (i.e. the linearisation method).

Im not sure what to do i made $f(x) = e^x$ and $f'(x) = e^x$ but I'm not sure what i would chose as my $x_0$ and my $x_1$ and why?

user639649
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Notice that $x=\ln(10)+0.1$ is just a little bit more than $\ln(10)$. So you should find the linearization of $f(x)=e^x$ at $x=\ln(10)$, and then evaluate this at $x=\ln(10)+0.1$.

kccu
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  • I was wondering what defines 'small' because my lecturer told me that the gap between the two x's should be 'small' and i thought that small meant between 0-0.99 so then it wouldnt classify – user639649 May 07 '19 at 01:09
  • Also I would think to put ln(10) - 0.1 in the calculator to find out its distance, (so making $x_0 = ln(10)$ and $x_1 = 0.1$ – user639649 May 07 '19 at 01:12
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    @user639649: in this context, small means small enough that the quadratic and higher terms in the Taylor expansion are negligible. There is no given value that is small. – Ross Millikan May 07 '19 at 01:14
  • so what would the
    $x_0$ and $x_1$ be in the case?     – user639649 May 07 '19 at 01:47
  • This should probably be a comment –  May 07 '19 at 01:50
  • @user639649 I don't know what "$x_0$" and "$x_1$" mean to you. Those don't have a standard meaning when talking about linearization. You should post the particular linearization formula you are using so we can see what $x_0$ and $x_1$ mean. – kccu May 07 '19 at 12:05