Question

If f(x) =root under x and phi(x) =1/root x in (a, b), then verify cauchy mean value theorem.

Answer 1

www.AssignmentExpert.com

TASK№1

If $f(x) = \sqrt{x}$ and $\phi(x) = \frac{1}{\sqrt{x}}$ in $(a, b)$, then verify cauchy mean value theorem.

SOLUTION

First of all point to the possible values of $a$ and $b$. Since the task is to verify the Cauchy theorem about average for the functions

$f(x) = \sqrt{x}$ and $\phi(x) = \frac{1}{\sqrt{x}}$ that are defined for $\forall x > 0$, so that $a > 0$ and $b > 0$. Without loss of generality we can assume that $a < b$. Recall the Cauchy's theorem about average:

Cauchy's mean value theorem, also known as the extended mean value theorem, is a generalization of the mean value theorem. It states: If functions $f(x)$ and $\phi(x)$ are both continuous on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, then there exists some $c \in (a,b)$, such that

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

To be convinced of the truth of the theorem is necessary to solve the equation with respect to $C$ and show that it is $c \in (a, b)$

$$\begin{pmatrix} f(x) = \sqrt{x} & \longrightarrow & f'(x) = \frac{1}{2\sqrt{x}} \\ \phi(x) = \frac{1}{\sqrt{x}} & \longrightarrow & \phi'(x) = -\frac{1}{2} \frac{1}{\sqrt{x^3}} \end{pmatrix}$$ $$\begin{aligned} \frac{f'(c)}{\phi'(c)} &= \frac{f(a) - f(b)}{\phi(a) - \phi(b)} \longrightarrow \frac{\frac{1}{2\sqrt{c}}}{-\frac{1}{2} \frac{1}{\sqrt{c^3}}} = \frac{\sqrt{a} - \sqrt{b}}{\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}}} \\ -c &= \frac{\sqrt{a} - \sqrt{b}}{\sqrt{b} - \sqrt{a}} \longrightarrow c = \sqrt{ab} \end{aligned}$$

We show that $c > a$

$$ab > a^2 \longrightarrow a(b - a) > 0 - \text{because we have agreed that } a > 0 \text{ and } a < b$$

We show that $c < b$

$$ab < b^2 \longrightarrow b(a - b) < 0 - \text{because we have agreed that } b > 0 \text{ and } a < b$$

We see that the found value $c$ really lies in the interval $(a, b)$

Cauchy's mean value theorem verified.