Answer
(b) $2=\max \Delta x_{k}$
$(\text { a }) \sum_{k=1}^{4} f\left(x_{k}^{*}\right) \Delta x_{k}^{*}=0$
Work Step by Step
a.
We find:
$\sum_{k=1}^{4} \Delta x_{k}^{*} f\left(x_{k}^{*}\right) = \Delta x_{1}^{*} f\left(x_{1}^{*}\right)+\Delta x_{2}^{*} f\left(x_{2}^{*}\right) + \Delta x_{3}^{*} f\left(x_{3}^{*}\right)+\Delta x_{4}^{*} f\left(x_{4}^{*}\right) $
$=2\cdot f(-2) +1\cdot f(0) +1\cdot f(0) +2 \cdot f(2) $
$=2 \cdot -8 +1 \cdot 0+1 \cdot 0+2 \cdot 8$
$=0$
(b) $\max \Delta x_{k}=\max \{1,1,2,2\}=2$
