A function $f$ has derivatives of all orders for all $x$. Some values of $f^{ }′$ are given in the table below. It is known that $f(2) = -5$.

1. Use a trapezoidal sum with four subintervals to approximate $\int _ { 2 } ^ { 2.8 } f ^ { \prime } ( x ) d x$.

   Hint (a):

   $0.2 ( \frac { 1 } { 2 } ) ( 4 + 2 ( 12 ) + 2 ( 23 ) + 2 ( 39 ) + 59$

2. Use your approximation from part (a) to estimate the value of $f(2.8)$. Justify your estimate with your work.

   Hint (b):

   $f ( 2.8 ) = f ( 2 ) + \int _ { 2 } ^ { 2.8 } f ^ { \prime } ( x ) d x$

3. Use Euler’s Method, starting with $x = 2$, with four steps of equal size, to approximate $f(2.8)$. Show your work.

   Hint (c):

   At $x=2$, the slope is $4$. Since the step size is $dx=.02$, then $\frac { d y } { d x } = \frac { 4 } { 1 } = \frac { 4 ( 0.2 ) } { 0.2 }$

   Starting at $(2,-5)$, the next point is: $(2 + 0.2 , - 5 + 4 ( 0.2 ))$

4. If $f^{\prime\prime}(2) = 32$ and $f^{\prime\prime\prime}(2) = 72$, write a third-degree Taylor polynomial about $x = 2$ and use it to approximate $f(2.8)$.

   Hint (d):

   $p _ { 3 } ( x ) = f ( 2 ) + f ^ { \prime } ( 2 ) ( x - 2 ) + \frac { f ^ { \prime \prime } ( 2 ) } { 2 ! } ( x - 2 ) ^ { 2 } + \frac { f ^ { \prime \prime \prime } ( 2 ) } { 3 ! } ( x - 2 ) ^ { 3 }$

5. How close were your estimates of $f(2.8)$ in parts (b) through (d)? Explain why this happened.