a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512…

a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512. 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(-512,512) and is differentiable on -512,512. 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 (-512,512). c. the mean value theorem does not apply to the given function because f(x) is not continuous on -512,512. d. f(x) is continuous on -512,512 and is differentiable on (-512,512). therefore, the mean value theorem applies to the given function. 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.

a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512. 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(-512,512) and is differentiable on -512,512. 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 (-512,512). c. the mean value theorem does not apply to the given function because f(x) is not continuous on -512,512. d. f(x) is continuous on -512,512 and is differentiable on (-512,512). therefore, the mean value theorem applies to the given function. 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.

Answer

Explanation:

Step1: Check continuity and differentiability

The function $f(x)=6x^{\frac{1}{9}}$ is continuous on the closed - interval $[-512,512]$ and differentiable on the open - interval $(-512,512)$. 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 number $c\in(a,b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}$. Here $a=-512$, $b = 512$.

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

First, find $f(-512)$ and $f(512)$. $f(-512)=6(-512)^{\frac{1}{9}}=6(-2)=-12$ (since $(-2)^9=-512$). $f(512)=6(512)^{\frac{1}{9}}=6\times2 = 12$ (since $2^9 = 512$). Then $\frac{f(512)-f(-512)}{512-(-512)}=\frac{12-(-12)}{512 + 512}=\frac{24}{1024}=\frac{3}{128}$.

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

The derivative of $f(x)=6x^{\frac{1}{9}}$ using the power rule $(x^n)^\prime=nx^{n - 1}$ is $f^{\prime}(x)=6\times\frac{1}{9}x^{\frac{1}{9}-1}=\frac{2}{3}x^{-\frac{8}{9}}$.

Step4: Solve for $x$

Set $f^{\prime}(x)=\frac{3}{128}$, so $\frac{2}{3}x^{-\frac{8}{9}}=\frac{3}{128}$. Cross - multiply to get $2\times128 = 3\times3x^{\frac{8}{9}}$. $256 = 9x^{\frac{8}{9}}$. $x^{\frac{8}{9}}=\frac{256}{9}$. Raise both sides to the $\frac{9}{8}$ power: $x=\pm(\frac{256}{9})^{\frac{9}{8}}=\pm\frac{2^{9}}{3^{\frac{9}{8}}}=\pm\frac{512}{3^{\frac{9}{8}}}$. Since $x\in(-512,512)$, the values of $x$ are valid.

Answer:

a. D. $f(x)$ is continuous on $[-512,512]$ and is differentiable on $(-512,512)$. Therefore, the Mean Value Theorem applies to the given function. b. A. The point(s) is/are $x=\pm\frac{512}{3^{\frac{9}{8}}}$