Answer:
[tex]x \approx 3.195[/tex] satisfies the conclusion of the Mean Value Theorem for [tex]f(x) = \ln x^{3}[/tex] over the interval [tex][1,e^{2}][/tex].
Step-by-step explanation:
According to the Mean Value Theorem, for all function that is differentiable over the interval [tex][a, b][/tex], there is at a value [tex]c[/tex] within the interval such that:
[tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex] (1)
Where:
[tex]a[/tex], [tex]b[/tex] - Lower and upper bounds.
[tex]f(a)[/tex], [tex]f(b)[/tex] - Function evaluated at lower and upper bounds.
[tex]f'(c)[/tex] - First derivative of the function evaluated at [tex]c[/tex].
If we know that [tex]f(x) = \ln x^{3} = 3\cdot \ln x[/tex], [tex]f'(x) = \frac{3}{x}[/tex], [tex]a = 1[/tex] and [tex]b = e^{2}[/tex], then we find that:
[tex]\frac{3}{c} = \frac{3\cdot \ln e^{2}-3\cdot \ln 1}{e^{2}-1}[/tex]
[tex]\frac{3}{c} = \frac{6\cdot \ln e-3\cdot \ln 1 }{e^{2}-1 }[/tex]
[tex]\frac{3}{c} = \frac{6}{e^{2}-1}[/tex]
[tex]c = \frac{1}{2}\cdot (e^{2}-1)[/tex]
[tex]c \approx 3.195[/tex]
[tex]x \approx 3.195[/tex] satisfies the conclusion of the Mean Value Theorem for [tex]f(x) = \ln x^{3}[/tex] over the interval [tex][1,e^{2}][/tex].
