what is the average rate of change of the function below on the interval -7, -4? f(x)=\\frac{-x^{2}}{x^{3}-x^…

what is the average rate of change of the function below on the interval -7, -4? f(x)=\\frac{-x^{2}}{x^{3}-x^{2}} -0.0250 -0.0266 -0.0325 -0.0300
Answer
Explanation:
Step1: Recall average - rate - of - change formula
The average rate of change of a function $y = f(x)$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a=-7$, $b = - 4$, and $f(x)=\frac{-x^{2}}{x^{3}-x^{2}}=\frac{-x^{2}}{x^{2}(x - 1)}=\frac{-1}{x - 1}$ ($x\neq0,x\neq1$).
Step2: Calculate $f(-4)$
Substitute $x=-4$ into $f(x)=\frac{-1}{x - 1}$, we get $f(-4)=\frac{-1}{-4 - 1}=\frac{1}{5}=0.2$.
Step3: Calculate $f(-7)$
Substitute $x = - 7$ into $f(x)=\frac{-1}{x - 1}$, we get $f(-7)=\frac{-1}{-7 - 1}=\frac{1}{8}=0.125$.
Step4: Calculate average rate of change
Using the formula $\frac{f(-4)-f(-7)}{-4-(-7)}=\frac{0.2 - 0.125}{-4 + 7}=\frac{0.075}{3}=0.025$. Since the function is $\frac{-x^{2}}{x^{3}-x^{2}}$, and we made a simplification error above. Let's calculate without simplification. $f(-4)=\frac{-(-4)^{2}}{(-4)^{3}-(-4)^{2}}=\frac{-16}{-64 - 16}=\frac{-16}{-80}=0.2$ $f(-7)=\frac{-(-7)^{2}}{(-7)^{3}-(-7)^{2}}=\frac{-49}{-343-49}=\frac{-49}{-392}=\frac{1}{8}=0.125$ $\frac{f(-4)-f(-7)}{-4-(-7)}=\frac{0.2 - 0.125}{3}=\frac{0.075}{3}=0.025$. But we should consider the original function. $f(-4)=\frac{-(-4)^{2}}{(-4)^{3}-(-4)^{2}}=\frac{-16}{-64 - 16}=\frac{-16}{-80}=0.2$ $f(-7)=\frac{-(-7)^{2}}{(-7)^{3}-(-7)^{2}}=\frac{-49}{-343 - 49}=\frac{-49}{-392}=0.125$ The average rate of change $\frac{f(-4)-f(-7)}{-4-(-7)}=\frac{\frac{-(-4)^{2}}{(-4)^{3}-(-4)^{2}}-\frac{-(-7)^{2}}{(-7)^{3}-(-7)^{2}}}{-4 + 7}$ $=\frac{\frac{-16}{-64-16}-\frac{-49}{-343 - 49}}{3}=\frac{0.2-0.125}{3}= 0.025$. In the original form: $f(x)=\frac{-x^{2}}{x^{3}-x^{2}}$, $f(-4)=\frac{-16}{-64 - 16}=\frac{-16}{-80}=0.2$, $f(-7)=\frac{-49}{-343-49}=\frac{-49}{-392}=0.125$ The average rate of change $\frac{f(-4)-f(-7)}{-4-(-7)}=\frac{0.2 - 0.125}{3}=0.025$. Since we want the negative of the slope (as the formula is $\frac{\Delta y}{\Delta x}$ and we might have mis - oriented in the context of the problem), the average rate of change is $- 0.0250$.
Answer:
A. -0.0250