Question

What value of x satisfies the conclusion of the mean value theorem for f(x) = ln(x3) over the interval [1, e2]?

Answer 1

Answer:

$x \approx 3.195$ satisfies the conclusion of the Mean Value Theorem for $f(x) = \ln x^{3}$ over the interval $[1,e^{2}]$.

Step-by-step explanation:

According to the Mean Value Theorem, for all function that is differentiable over the interval $[a, b]$, there is at a value $c$ within the interval such that:

$f'(c) = \frac{f(b)-f(a)}{b-a}$ (1)

Where:

$a$, $b$ - Lower and upper bounds.

$f(a)$, $f(b)$ - Function evaluated at lower and upper bounds.

$f'(c)$ - First derivative of the function evaluated at $c$.

If we know that $f(x) = \ln x^{3} = 3\cdot \ln x$, $f'(x) = \frac{3}{x}$, $a = 1$ and $b = e^{2}$, then we find that:

$\frac{3}{c} = \frac{3\cdot \ln e^{2}-3\cdot \ln 1}{e^{2}-1}$

$\frac{3}{c} = \frac{6\cdot \ln e-3\cdot \ln 1 }{e^{2}-1 }$

$\frac{3}{c} = \frac{6}{e^{2}-1}$

$c = \frac{1}{2}\cdot (e^{2}-1)$

$c \approx 3.195$

$x \approx 3.195$ satisfies the conclusion of the Mean Value Theorem for $f(x) = \ln x^{3}$ over the interval $[1,e^{2}]$.