Question

a. determine whether the mean value theorem applies to the function ( f(x)=6x^{3} ) on the interval (-8,8).
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. ( f(x) ) is continuous on ((-8,8)) and is differentiable on (-8,8). therefore, the mean value theorem applies to the given function
b. the mean value theorem does not apply to the given function because ( f(x) ) is not differentiable on ((-8,8))
c. ( f(x) ) is continuous on (-8,8) and is differentiable on ((-8,8)). therefore, the mean value theorem applies to the given function
d. the mean value theorem does not apply to the given function because ( f(x) ) is not continuous on (-8,8).
b. select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. the point(s) is/are ( x=)
(type exponential notation with positive exponents. use integers or fractions for any numbers in the expression.)
b. the mean value theorem does not apply in this case

Explanation:

Step1: Check continuity and differentiability

A polynomial function \(y = ax^{n}+bx^{n - 1}+\cdots+c\) is continuous and differentiable everywhere. The function \(f(x)=6x^{3}\) is a polynomial. Polynomial functions are continuous on the closed interval \([-8,8]\) (by the property of polynomial continuity: \(\lim_{x
ightarrow a}P(x)=P(a)\) for all \(a\in\mathbb{R}\) and \(P(x)\) polynomial) and differentiable on the open interval \((-8,8)\) (since \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{6(x + h)^{3}-6x^{3}}{h}=\lim_{h
ightarrow0}\frac{6(x^{3}+3x^{2}h + 3xh^{2}+h^{3})-6x^{3}}{h}=\lim_{h
ightarrow0}(18x^{2}+18xh + 6h^{2})=18x^{2}\), which exists for all \(x\in(-8,8)\)).

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 \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Here, \(a=-8\), \(b = 8\), \(f(x)=6x^{3}\), \(f(-8)=6\times(-8)^{3}=6\times(-512)=-3072\), \(f(8)=6\times8^{3}=6\times512 = 3072\). Then \(\frac{f(8)-f(-8)}{8-(-8)}=\frac{3072-(-3072)}{16}=\frac{6144}{16}=384\). And \(f^{\prime}(x)=18x^{2}\). Set \(f^{\prime}(c)=18c^{2}\) equal to \(\frac{f(8)-f(-8)}{8 - (-8)}\). So \(18c^{2}=384\), \(c^{2}=\frac{384}{18}=\frac{64}{3}\), \(c=\pm\frac{8}{\sqrt{3}}=\pm\frac{8\sqrt{3}}{3}\)

Answer:

a. C. \(f(x)\) is continuous on \([-8,8]\) and is differentiable on \((-8,8)\). Therefore, the Mean Value Theorem applies to the given function
b. A. The point(s) is/are \(x =-\frac{8\sqrt{3}}{3},\frac{8\sqrt{3}}{3}\)