a. determine whether the mean value theorem applies to the function f(x)=x + 1/x on the interval -4,-3. b…

Explanation:

Step1: Check continuity

The function $f(x)=x + \frac{1}{x}$ is a sum of a polynomial $y = x$ and a rational - function $y=\frac{1}{x}$. The rational function $y=\frac{1}{x}$ is undefined at $x = 0$, but the interval $[-4,-3]$ does not contain $x = 0$. A polynomial is continuous everywhere and a rational function $\frac{1}{x}$ is continuous on its domain. So, $f(x)$ is continuous on $[-4,-3]$.

Step2: Check differentiability

Differentiate $f(x)$ using the sum - rule and the power - rule. The derivative of $y=x$ is $y^\prime=1$, and the derivative of $y = \frac{1}{x}=x^{-1}$ is $y^\prime=-x^{-2}=-\frac{1}{x^{2}}$. So, $f^\prime(x)=1-\frac{1}{x^{2}}$. This derivative exists for all $x\neq0$. Since $x\in(-4,-3)$ (the open - interval corresponding to $[-4,-3]$) and $0\notin(-4,-3)$, the function $f(x)$ is differentiable on $(-4,-3)$.

Step3: Apply Mean - Value Theorem

The Mean - Value Theorem states that if $y = f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c\in(a,b)$ such that $f^\prime(c)=\frac{f(b)-f(a)}{b - a}$. Here, $a=-4$, $b = - 3$, $f(a)=-4-\frac{1}{4}=-\frac{16 + 1}{4}=-\frac{17}{4}$, $f(b)=-3-\frac{1}{3}=-\frac{9 + 1}{3}=-\frac{10}{3}$. Then $\frac{f(b)-f(a)}{b - a}=\frac{-\frac{10}{3}+\frac{17}{4}}{-3 + 4}=\frac{-\frac{40}{12}+\frac{51}{12}}{1}=\frac{11}{12}$. Set $f^\prime(c)=1-\frac{1}{c^{2}}=\frac{11}{12}$. Solve for $c$: [ \begin{align*} 1-\frac{1}{c^{2}}&=\frac{11}{12}\ -\frac{1}{c^{2}}&=\frac{11}{12}-1\ -\frac{1}{c^{2}}&=\frac{11 - 12}{12}\ -\frac{1}{c^{2}}&=-\frac{1}{12}\ c^{2}&=12\ c&=\pm2\sqrt{3} \end{align*} ] Since $c\in(-4,-3)$, $c=-2\sqrt{3}\approx - 3.46$.

Answer:

a. C. Yes, because the function is continuous on the interval $[-4,-3]$ and differentiable on the interval $(-4,-3)$. b. $c=-2\sqrt{3}\approx - 3.46$