select value of the twice differentiable function f and its derivative are shown in the table below. what is…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = g(x)$ over the interval $[a,b]$ is given by $\frac{1}{b - a}\int_{a}^{b}g(x)dx$. For a discrete - case (using data from a table), we can use the formula $\frac{\sum_{i = 1}^{n}g(x_i)\Delta x_i}{\sum_{i = 1}^{n}\Delta x_i}$. Here, $a = 2$, $b = 7$, and we can consider the sub - intervals. The sub - intervals are $[2,5]$, $[5,7]$. The lengths of the sub - intervals are $\Delta x_1=5 - 2 = 3$ and $\Delta x_2=7 - 5 = 2$.

Step2: Calculate the sum of $f'(x)\Delta x$

For the sub - interval $[2,5]$, $f'(x)=3$ and $\Delta x = 3$, so $f'(x)\Delta x=3\times3 = 9$. For the sub - interval $[5,7]$, $f'(x)=0$ and $\Delta x = 2$, so $f'(x)\Delta x=0\times2 = 0$. The sum $\sum_{i}f'(x_i)\Delta x_i=9 + 0=9$.

Step3: Calculate the sum of $\Delta x$

The sum of the lengths of the sub - intervals $\sum_{i}\Delta x_i=(5 - 2)+(7 - 5)=3 + 2 = 5$.

Step4: Calculate the average value

The average value of $f'$ from $x = 2$ to $x = 7$ is $\frac{\sum_{i}f'(x_i)\Delta x_i}{\sum_{i}\Delta x_i}=\frac{9}{5}= \frac{1}{5}\times9$. But we can also use the formula $\frac{f'(x_1)+f'(x_2)}{2}$ (a simpler way when the sub - intervals are of equal length, which is not the case here, but we can still use the general formula). Another way is to use the fact that the average value of a function $y = f'(x)$ over $[a,b]$ is $\frac{1}{b - a}\int_{a}^{b}f'(x)dx=\frac{f(b)-f(a)}{b - a}$ (by the fundamental theorem of calculus). Here, $a = 2$, $b = 7$, $f(7)=11$, $f(2)=1$. First, we know that $\int_{2}^{7}f'(x)dx=f(7)-f(2)=11 - 1 = 10$. The average value of $f'$ over $[2,7]$ is $\frac{1}{7 - 2}\int_{2}^{7}f'(x)dx=\frac{10}{5}=2$.

Answer:

2