2. use the mean value theorem, if it applies, to find an x - value where the instantaneous slope there is…
2. use the mean value theorem, if it applies, to find an x - value where the instantaneous slope there is the same as the average slope over the given interval. f(x) = (x² - 9)/(x + 3) -5,2 cannot be determined ∞ 0 -3
Answer
Explanation:
Step1: Simplify the function
First, simplify $f(x)=\frac{x^{2}-9}{x + 3}$. Since $x^{2}-9=(x + 3)(x - 3)$, then $f(x)=\frac{(x + 3)(x - 3)}{x+3}=x - 3$ for $x\neq - 3$. The function $y=x - 3$ is a linear - function and is continuous on the closed interval $[-5,2]$ and differentiable on the open interval $(-5,2)$.
Step2: Calculate the average slope
The average slope of the function $y = f(x)$ over the interval $[a,b]=[-5,2]$ is given by $\frac{f(b)-f(a)}{b - a}$. Here, $a=-5$, $b = 2$, $f(a)=f(-5)=-5 - 3=-8$, $f(b)=f(2)=2 - 3=-1$. So the average slope is $\frac{f(2)-f(-5)}{2-(-5)}=\frac{-1-(-8)}{2 + 5}=\frac{-1 + 8}{7}=1$.
Step3: Calculate the derivative
The derivative of $f(x)=x - 3$ is $f^\prime(x)=1$. Since $f^\prime(x)$ is a constant function and $f^\prime(x)=1$ for all $x$ in the domain of $f(x)$ (the domain of the simplified function $y=x - 3$ in the context of the original problem is all real numbers except $x=-3$, and $(-5,2)$ is part of this domain), for any $x\in(-5,2)$, the instantaneous slope is equal to the average slope. We can choose $x = 0$ which is in the open interval $(-5,2)$.
Answer:
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