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Using the points (0,2), (1,4), and (2,8) suppose $f(x)=2^{(x+1)}$ is an approximation. What is the maximum error on [0,3]?

I know the error formula is $$\frac{f'''(\varepsilon(x))}{(n+1)!}(x-x_0)(x-x_1)(x-x_2)$$

so evaluating what I have I got $$\frac{2^{x+1}log^3(2)}{6}x^3-3x^2+2x$$

I am stuck at this point I am not sure if I should simplify or evaluate for the interval. Thanks in advance.

ECollins
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  • Do you know how to maximize a smooth function (such as your error)? – anderstood Sep 22 '16 at 00:23
  • Didn't you forget to derive $f$ in your evaluation? And parentheses. – anderstood Sep 22 '16 at 00:36
  • I am unfamiliar with maximizing smooth functions we are just learning about lagrange interpolation. as for deriving $f$ i did the triple derivative of $f(x)$ and got the numerator in my evaluation @anderstood – ECollins Sep 22 '16 at 00:44
  • OK. I don't understand what the error refers to: error of what? If $n=3$, why is your denominator $6$ and not $4!=24$ (maybe it simplified, I did not do the calculation)? I would say the error refers to the distance between the points $(x_i,y_i)$ and the points $(x_i,f(x_i))$. The simplest is probably to compute the three errors. – anderstood Sep 22 '16 at 00:49

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