a. determine whether the mean value theorem applies to the function f(x)=x + 1/x on the interval 3,7 b. if…

a. determine whether the mean value theorem applies to the function f(x)=x + 1/x on the interval 3,7 b. if so, find or approximate the point(s) that are guaranteed to exist by the mean value theorem. a. choose the correct answer below. a. no, because the function is not continuous on the interval 3,7, and is not differentiable on the interval (3,7). b. yes, because the function is continuous on the interval 3,7 and differentiable on the interval (3,7). c. no, because the function is differentiable on the interval (3,7), but is not continuous on the interval 3,7. d. no, because the function is continuous on the interval 3,7, but is not differentiable on the interval (3,7).

a. determine whether the mean value theorem applies to the function f(x)=x + 1/x on the interval 3,7 b. if so, find or approximate the point(s) that are guaranteed to exist by the mean value theorem. a. choose the correct answer below. a. no, because the function is not continuous on the interval 3,7, and is not differentiable on the interval (3,7). b. yes, because the function is continuous on the interval 3,7 and differentiable on the interval (3,7). c. no, because the function is differentiable on the interval (3,7), but is not continuous on the interval 3,7. d. no, because the function is continuous on the interval 3,7, but is not differentiable on the interval (3,7).

Answer

Explanation:

Step1: Check continuity and differentiability

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 for the interval ([3,7]), it is continuous and differentiable. The sum of two continuous and differentiable functions on an interval is also continuous and differentiable on that interval. A function (y = f(x)) is continuous on ([a,b]) and differentiable on ((a,b)) to satisfy the Mean - Value Theorem. Here, (a = 3), (b = 7), and (f(x)=x+\frac{1}{x}) is continuous on ([3,7]) and differentiable on ((3,7)) since (f^\prime(x)=1-\frac{1}{x^{2}}) exists for all (x\in(3,7)).

Step2: Apply the Mean - Value Theorem formula

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}). First, find (f(3)) and (f(7)): [ \begin{align*} f(3)&=3+\frac{1}{3}=\frac{9 + 1}{3}=\frac{10}{3}\ f(7)&=7+\frac{1}{7}=\frac{49+1}{7}=\frac{50}{7} \end{align*} ] Then (\frac{f(7)-f(3)}{7 - 3}=\frac{\frac{50}{7}-\frac{10}{3}}{4}=\frac{\frac{150 - 70}{21}}{4}=\frac{\frac{80}{21}}{4}=\frac{20}{21}). Next, find (f^\prime(x)=1-\frac{1}{x^{2}}), and set (f^\prime(c)=\frac{20}{21}). So (1-\frac{1}{c^{2}}=\frac{20}{21}). [ \begin{align*} 1-\frac{20}{21}&=\frac{1}{c^{2}}\ \frac{1}{21}&=\frac{1}{c^{2}}\ c^{2}&=21\ c&=\sqrt{21}\approx4.58 \end{align*} ] Since (3\lt\sqrt{21}\lt7), (c = \sqrt{21}) is in the interval ((3,7)).

Answer:

a. B. Yes, because the function is continuous on the interval ([3,7]) and differentiable on the interval ((3,7)) b. (c=\sqrt{21}\approx4.58)