5. use the mean value theorem, if it applies, to find an x - value where the instantaneous slope there is…

Explanation:

Step1: Recall Mean - Value Theorem

The Mean - Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and differentiable on the open interval ((a,b)), then there exists at least one (c\in(a,b)) such that (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}). Here, (a=-10), (b = 3), and (f(x)=x^{2}+6x - 22).

Step2: Calculate (f(a)) and (f(b))

First, find (f(-10)): [ \begin{align*} f(-10)&=(-10)^{2}+6\times(-10)-22\ &=100-60 - 22\ &=18 \end{align*} ] Next, find (f(3)): [ \begin{align*} f(3)&=3^{2}+6\times3-22\ &=9 + 18-22\ &=5 \end{align*} ] Then, calculate (\frac{f(3)-f(-10)}{3-(-10)}): [ \begin{align*} \frac{f(3)-f(-10)}{3-(-10)}&=\frac{5 - 18}{3 + 10}\ &=\frac{-13}{13}\ &=-1 \end{align*} ]

Step3: Find the derivative of (f(x))

Differentiate (f(x)=x^{2}+6x - 22) with respect to (x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we get (f^{\prime}(x)=2x+6).

Step4: Set (f^{\prime}(x)) equal to the average slope

Set (f^{\prime}(x)=-1), so (2x+6=-1). Solve for (x): [ \begin{align*} 2x&=-1 - 6\ 2x&=-7\ x&=-\frac{7}{2} \end{align*} ]

Answer:

(-\frac{7}{2})