points: 20 points\n1. (3 points) the graph of f(x) is shown below. list the following quantities in order…

Explanation:

Step1: Recall rate - of - change formulas

The average rate of change of $y = f(x)$ over $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$, and the instantaneous rate of change at $x = c$ is $f^{\prime}(c)$ (the slope of the tangent line at $x = c$).

Step2: Analyze average rate of change $A$

For the average rate of change $A$ between $x = 1$ and $x = 3$, $A=\frac{f(3)-f(1)}{3 - 1}$. From the graph, the secant line between $x = 1$ and $x = 3$ has a positive slope.

Step3: Analyze average rate of change $B$

For the average rate of change $B$ between $x = 5$ and $x = 6$, $B=\frac{f(6)-f(5)}{6 - 5}=f(6)-f(5)$. From the graph, the secant line between $x = 5$ and $x = 6$ has a negative slope.

Step4: Analyze instantaneous rate of change $C$

The instantaneous rate of change $C = f^{\prime}(1)$. The tangent line at $x = 1$ has a positive slope, and it is steeper than the secant line between $x = 1$ and $x = 3$, so $C>A$.

Step5: Analyze instantaneous rate of change $D$

The instantaneous rate of change $D = f^{\prime}(3)$. The tangent line at $x = 3$ has a positive slope, but it is less steep than the tangent line at $x = 1$, so $A < D<C$.

Step6: Analyze instantaneous rate of change $E$

The instantaneous rate of change $E = f^{\prime}(5)$. The tangent line at $x = 5$ has a negative slope, but its magnitude is less than the magnitude of the slope of the secant line between $x = 5$ and $x = 6$.

Step7: Analyze instantaneous rate of change $F$

The instantaneous rate of change $F = f^{\prime}(6)$. The tangent line at $x = 6$ has a negative slope, and it is steeper (more negative) than the tangent line at $x = 5$ and the secant line between $x = 5$ and $x = 6$.

So, $F<B<E<A<D<C$.

Answer:

$F < B < E < A < D < C$