the mean value theorem can be applied to which of the following functions on the closed interval $-5,5$?\na…

Explanation:

Step1: Recall the conditions for the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function (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}).

Step2: Analyze function (A): (f(x)=\frac{1}{\sin x})

The function (y = \sin x) has a zero at (x = 0\in[- 5,5]). So (f(x)=\frac{1}{\sin x}) is not continuous on ([-5,5]) (since (\lim_{x\rightarrow0}\frac{1}{\sin x}) does not exist in the sense of finite - valued limits).

Step3: Analyze function (B): (f(x)=\frac{x - 1}{|x - 1|})

We can rewrite (f(x)) as a piece - wise function. (f(x)=\left{\begin{array}{ll}1, & x>1\- 1, & x<1\end{array}\right.). The function (f(x)) has a jump discontinuity at (x = 1\in[-5,5]). So (f(x)) is not continuous on ([-5,5]).

Step4: Analyze function (C): (f(x)=\frac{x^{2}}{x^{2}-36}=\frac{x^{2}}{(x + 6)(x - 6)})

The denominator (x^{2}-36=(x + 6)(x - 6)). The function (f(x)) is undefined when (x=\pm6). Since (\pm6\notin[-5,5]), the function (y = f(x)) is a rational function and is continuous on ([-5,5]) (because the denominator is non - zero on ([-5,5])). The derivative (f^{\prime}(x)=\frac{2x(x^{2}-36)-x^{2}(2x)}{(x^{2}-36)^{2}}=\frac{-72x}{(x^{2}-36)^{2}}) exists for all (x\in(-5,5)) (denominator ((x^{2}-36)^{2}>0) for (x\in(-5,5))).

Step5: Analyze function (D): (f(x)=\frac{x^{2}}{x^{2}-4}=\frac{x^{2}}{(x + 2)(x - 2)})

The denominator (x^{2}-4=(x + 2)(x - 2)). The function (f(x)) is undefined at (x=\pm2) and (x=\pm2\in[-5,5]). So (f(x)) is not continuous on ([-5,5]).

Answer:

C. (f(x)=\frac{x^{2}}{x^{2}-36})