![Which function has the greater average rate over the interval [-2,-1]?x | f(x)---------2 | 10-1 | 80 | 61 | 4[tex]g(x) = {x}^{2} - 2x + 1[/tex] class=](https://us-static.z-dn.net/files/de0/f7feeb77b9a9e088c33889ba5bc22f14.png)
Answer :
First function,
F(x)
The find the greatest rate of change, you will find the slope over the given interval.
The slope of the function f(x) is m
[tex]m\text{ = }\frac{y_2-y_1_{}_{}_{}}{x_2-x_1}[/tex]From the table,
x1 = -2, y1 = 10
x2 = -1, y2 = 8
[tex]\begin{gathered} m\text{ = }\frac{8\text{ - 10}}{-1\text{ - (-2)}} \\ m\text{ = }\frac{-2}{-1\text{ + 2}} \\ m\text{ = }\frac{-2}{1} \\ m\text{ = -2} \end{gathered}[/tex]Second, function g(x)
[tex]\begin{gathered} g(x)=-x^2\text{ - 2x + 1} \\ whenx_1=-2,y_{1\text{ }}=-(-2)^{2\text{ }}-\text{ 2(-2) + 1 = -4 + 4 + 1 = 1} \\ \text{when x}_2=-1,y_2=-(-1)^2\text{ - 2(-1) + 1 = -1 + 2 + 1 = 2} \\ \text{next, find the rate of change} \\ m\text{ = }\frac{2\text{ - 1}}{-1\text{ -(-2)}} \\ m\text{ = }\frac{1}{-1\text{ + 2}} \\ m\text{ = }\frac{1}{1} \\ m\text{ = 1} \end{gathered}[/tex]Third function h(x)
[tex]\begin{gathered} \text{From the graph of h(x)} \\ \text{when x}_1=-1interceptthecurveat0,therefore.y_1\text{ = 0} \\ \text{when x}_2\text{ = -2 intercept the curve at }3,therefore,y_2\text{ = 3} \\ m\text{ = }\frac{\text{3 - 0}}{-\text{ 1 -(-2)}} \\ m\text{ = }\frac{3}{-1\text{ + 2}} \\ m\text{ = }\frac{3}{1} \\ m\text{ = 3} \end{gathered}[/tex]From the solution above, the function h(x) has the greatest average rate over the interval (-2, -1).
Final answer
h(x) has the greatest overage rate over interval [-2, -1]