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**TshingY** Could someone check if this is the correct way to solve this problem.

For what n will the approximation f(x) be accurate to within .0000001 using

$\displaystyle E_{f} = \frac{(b-a)h^2}{6} f''(\mu)$, for some $\displaystyle \mu$ between a and b is.

$\displaystyle h= \frac{b-a}{n+2}$

I think $\displaystyle f''(\mu)$ for some $\displaystyle \mu$ between a and b means the absolute value of the max between a and b.

$\displaystyle f(x) = \int_0^1(\frac{1}{12}x^4 - \frac{1}{6}x^3)dx$

$\displaystyle .0000001 \leq \frac{(1-0)}{6} (\frac{1-0}{n+2})^2 f''(\mu)$

$\displaystyle f''(\mu) = \mu^2 - \mu$

$\displaystyle .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * (\mu^2 - \mu)$

From this part on am I suppose to maximize $\displaystyle \mid f''(\mu)\mid$ on [0,1] and put that in place of $\displaystyle f''(\mu)$ and then solve for n?

If I do that I get

$\displaystyle .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * 5$, where 5 is the max of $\displaystyle \mid f''(\mu)\mid$ on [0,1]

Solving this for n gives 2886.75 and rounding it to 2887. So n would have to be 2887.