what is the instantaneous rate of change of the function below at x = - 2 with a step size of δx = 0.1…

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function $y = f(x)$ over the interval $[x,x+\Delta x]$ is $\frac{f(x + \Delta x)-f(x)}{\Delta x}$. Here, $x=-2$ and $\Delta x = 0.1$. First, find $f(x)$ at $x=-2$: [ \begin{align*} f(-2)&=\frac{-2 - 7}{(-2)^2-2\times(-2)-35}\ &=\frac{-9}{4 + 4-35}\ &=\frac{-9}{-27}\ &=\frac{1}{3} \end{align*} ]

Step2: Find $f(x+\Delta x)$

$x+\Delta x=-2 + 0.1=-1.9$ [ \begin{align*} f(-1.9)&=\frac{-1.9-7}{(-1.9)^2-2\times(-1.9)-35}\ &=\frac{-8.9}{3.61+3.8 - 35}\ &=\frac{-8.9}{7.41-35}\ &=\frac{-8.9}{-27.59}\ &\approx0.3225 \end{align*} ]

Step3: Calculate the average rate of change

[ \begin{align*} \frac{f(-1.9)-f(-2)}{0.1}&=\frac{0.3225-\frac{1}{3}}{0.1}\ &=\frac{0.3225 - 0.3333}{0.1}\ &=\frac{- 0.0108}{0.1}\ &=-0.108 \end{align*} ]

Answer:

-0.108