(a) Differentiating, we find f′′(x)=−x2+7x−6. Since f′′(x) is quadratic, we have a pretty good idea of what it looks like.
- It factors as f(x)=−(x−6)(x−1), so its two roots are at x=6 and x=1.
- The “flat part" of the parabola is at x=3.5 (since this is exactly half way between x=1 and x=6; alternately, we can check that f′′′(3.5)=0).
- Since the coefficient of x2 is negative, f(x) is increasing from −∞ to 3.5, then decreasing from 3.5 to ∞.
Therefore, over the interval [1,6], the largest positive value of f′′(x) occurs when x=3.5, and this is f′′(3.5)=−(3.5−6)(3.5−1)=6.25.

So, we take M=6.25.
(b) We differentiate further to find f(4)(x)=−2. This is constant everywhere, so we take L=∣−2∣=2.