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I have a function, which is $f(x) = e^{1/x}$. I want to calculate the error bound for the Trapezoid Rule, which formula is:

$$|E|\leq K\frac{(a-b)^3}{12\cdot n^2}$$

where $|f''(x)|\leq K$. What's the value of $K$ for the above function? If I calculated $f''(x)$ correctly, it should be: $$f"(x) = \frac{e^{1/x}}{x^3}\left(\frac 1x + 2\right).$$

Please correct me if I'm wrong. The boundaries are $[1, 2]$.

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    Can you write down $|f''(x)|$ explicitly first?2012-09-11
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    @Sam Done! I've updated my question2012-09-11
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    Alright, now you have it explicitly. You want an upper bound on it's absolute value. What happens to the second derivative when $x\rightarrow 0$ and $x\rightarrow\pm\infty$?2012-09-11
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    it goes to infinite when `x -> 0`, and goes to 0 when `x-> infinite`2012-09-11
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    This is the time to consider what interval you are integrating the function over. Does it include 0? Notice btw that $e^{1/x}$ has no limit as $x\rightarrow 0$ since from the positive side, it's infinity but from the negative side it's 02012-09-11
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    the boundaries are `[1, 2]` so `a=1` and `b=2`.2012-09-11
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    @Sam So my maxvalue of `K` is `3*e`?2012-09-11
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    Yep, it's a decreasing function (a fact that can be checked from the derivative). Just a minor point, you actually want a "min" value of $K$ since it sounds like you want to find the least upper bound.2012-09-11

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If you look at $f''$ you should be able to convince yourself that it is decreasing over $[1,2]$. In that case, the maximum value is $f''(1)$, which is then $K$