Question

determine whether the mean value theorem can be applied to ( f ) on the closed interval (a,b). (select all that apply.)
( f(x)=9 x^{3}, quad1,2 )
( square ) yes, the mean value theorem can be applied.
( square ) no, ( f ) is not continuous on (a, b).
( square ) no, ( f ) is not differentiable on ( (a, b) ).
( square ) none of the above.
if the mean value theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=\frac{f(b)-f(a)}{b - a} ). (enter your answers as a comma-separated list. if the mean value theorem cannot be applied, enter na.)
( c= )

Explanation:

Step1: Check continuity and differentiability

A polynomial function \(f(x)=9x^{3}\) is continuous and differentiable for all real numbers. So on the interval \([1,2]\), \(f(x)\) is continuous on \([1,2]\) and differentiable on \((1,2)\). Thus, the Mean Value Theorem can be applied.

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

Given \(a = 1\), \(b=2\), \(f(a)=f(1)=9\times1^{3}=9\), \(f(b)=f(2)=9\times2^{3}=9\times8 = 72\)

Step3: Calculate \(\frac{f(b)-f(a)}{b - a}\)

\(\frac{f(b)-f(a)}{b - a}=\frac{72 - 9}{2-1}=\frac{63}{1}=63\)

Step4: Find \(f^{\prime}(x)\) and solve \(f^{\prime}(c)\)

\(f^{\prime}(x)=\frac{d}{dx}(9x^{3})=27x^{2}\). Set \(f^{\prime}(c)=27c^{2}\) equal to \(\frac{f(b)-f(a)}{b - a}=63\). So \(27c^{2}=63\), \(c^{2}=\frac{63}{27}=\frac{7}{3}\), \(c=\pm\sqrt{\frac{7}{3}}\). But since \(c\in(1,2)\), \(c = \sqrt{\frac{7}{3}}\approx1.53\)

Answer:

Yes, the Mean Value Theorem can be applied. \(c=\sqrt{\frac{7}{3}}\)