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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? –  Alex R. Sep 11 '12 at 3:37
    
@Sam Done! I've updated my question –  philippe Sep 11 '12 at 3:39
    
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 $x\rightarrow\pm\infty$? –  Alex R. Sep 11 '12 at 3:40
    
it goes to infinite when x -> 0, and goes to 0 when x-> infinite –  philippe Sep 11 '12 at 3:42
    
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 0 –  Alex R. Sep 11 '12 at 3:43
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up vote 2 down vote accepted

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 $K$

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