Question

Let f : [0, 4] → R be a thrice differentiable function in (0, 4) such that

f(0) = −1, f(1) = 2, f(3) = −2, f(4) = 4.

Let g(x) = f(x)f '(x)f''(x) + 4. Using the mean value theorem, find the minimum number of distinct roots of g'(x) = 0.

Answer 1

$f:[0,4]\rightarrow \R$ is a thrice differentiable function in $(0,4)$

Therefore $f(x),f'(x),f''(x)$ is continuous on $(0,4)$ .

Now $f(0)=-1,f(1)=2,f(3)=-2,f(4)=4$

$\therefore$ By Intermediate-value theorem, [As f is continuous on [0,4] ]

$\exists x_1\in (0,1),x_2\in (1,3),x_3\in (3,4)\\ \ni\;\;f(x_i)=0\forall i\in \{1,2,3\}$

Now as f is continuous and differentiable on (0,4), by mean-value theorem,

$\exists x_4\in (x_1,x_2) \;and \;x_5\in (x_2,x_3)\ni\\ f'(x_4)=0=f'(x_5)$

Again as $f'$ is continuous and differentiable on (0,4), by mean-value theorem,

$\exists x_6\in (x_4,x_5)\ni\\ f''(x_6)=\frac{f'(x_5-x_4)}{x_5-x_4}=0$

Now putting $x_i$'s $\forall i\in \{1,2,3,4,5,6\}$,

$g(x_i)=f(x_i)f'(x_i)f''(x_i)+4=0+4=4$ ...(1)

$\underline{For \; minimal \; case, \; (x_6=x_2)},$

Now as $f,f',f''$ is continuous and differentiable on (0,4),

$g(x)=f(x)f'(x)f''(x)+4$ is continuous and differentiable on (0,4).

$\therefore$ From the fact (1), By Mean-Value Theorem, [ Particularly by Rolle's Theorem ]

$\exists x_7\in (x_1,x_4),x_8\in (x_4,x_2),x_9\in (x_2,x_5),x_{10}\in (x_5,x_3)\ni\\ g'(x_7)=g'(x_8)=g'(x_9)=g'(x_{10})=0$

Therefore $g'$ has at least $4$ distinct roots.

So, Minimum Number of distinct roots of $g' \, is\, 4.$