what is the instantaneous rate of change of the function below at x = 8 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, $x = 8$ and $\Delta x=0.1$.

Step2: Calculate $f(x)$ at $x = 8$

First, find $f(8)$. Given $f(x)=\frac{1}{x^{3}-6x^{2}+5x}$, then $x^{3}-6x^{2}+5x=8^{3}-6\times8^{2}+5\times8=512 - 384+40 = 168$. So $f(8)=\frac{1}{168}$.

Step3: Calculate $f(x+\Delta x)$ at $x = 8$ and $\Delta x = 0.1$

$x+\Delta x=8 + 0.1=8.1$. Then $(x+\Delta x)^{3}-6(x+\Delta x)^{2}+5(x+\Delta x)=8.1^{3}-6\times8.1^{2}+5\times8.1=531.441-393.66 + 40.5=178.281$. So $f(8.1)=\frac{1}{178.281}$.

Step4: Calculate the average rate of change

The average rate of change $\frac{f(8.1)-f(8)}{0.1}=\frac{\frac{1}{178.281}-\frac{1}{168}}{0.1}=\frac{\frac{168 - 178.281}{168\times178.281}}{0.1}=\frac{- 10.281}{168\times178.281\times0.1}$. $168\times178.281\times0.1 = 168\times17.8281=2995.1208$. $\frac{-10.281}{2995.1208}\approx - 0.0034$.

Answer:

-0.0034