Advanced Mean Value Theorem - Part 3

By Intermediate Value Theorem, there exists $c \in (0,~1)$ such that $f(c)=0$ .

Let $g(x)={x}^{3}f(x)$ . Then $g(0)=g(c)=0.$ By Rolle's Theorem, there exists $\xi \in (0,~c)$ such that

$g'(\xi)={\xi}^{3}f'(\xi)+3{\xi}^{2}f(\xi)=0$

Meanwhile, since $\xi>0$ , $~\xi f'(\xi)+3f(\xi)=0$ . Therefore, there always exists $\xi \in (0,~1)$ such that $\xi f'(\xi)+3f(\xi)=0$ .

The answer is yes, always.

$PROOF:$

Let $f(x)$ be a continuous function over $[0,1]$ , differentiable over $(0,1)$ , and satifies:

$f(0)=1$ (i)

$f(1)=-1$ (ii)

By the Mean-Value Theorem, there exists a tangent line to $f(x)$ at some point $\xi \in (0,1)$ such that $f'(\xi)$ equals the slope of the line connecting the two endpoints (i) and (ii). This slope is just equal to:

$f'(\xi) = \frac{-1-1}{1-0} =-2$

which when substituted into the above ODE yields:

$-2\xi + 3f(\xi) = 0 \Rightarrow f(\xi) = \frac{2}{3} \xi$ (iii)

and knowing that $-1 < f(\xi) < 1$ , we obtain $-\frac{3}{2} < \xi < \frac{3}{2}$ . Finally, we observe that $(0,1) \subset (-3/2, 3/2) \Rightarrow \boxed{\exists \xi \in (0,1)}$ that always satisfies the above ODE.

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$