Minimum number of roots 1

Calculus Level 5

Let $f(x)$ be a thrice differentiable function satisfying:

$|f(x) - f(4-x)| + |f(4-x)-f(4+x)| = 0, \forall x \in R$

If $f'(1)=0$ , then find the minimum number of roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$ , on $x \in [0,6]$ .

1 solution

Ashish Gupta
Apr 12, 2016

Since $f(x) = f(4-x) = f(4+x)$ , we have $f'(x) = -f'(4-x) = f'(4+x)$ for all $x \in \mathbb{R}$

Since $f'(1) = 0$ , using the above identity for $x = 1$ gives us $f'(1) = f'(3) = f'(5) = 0$ .

Also, using the above identity for $x = 2$ yields $f'(2) = -f'(2) = f'(6)$ , so $f'(2) = f'(6) = 0$ .

Similarly, for $x = 0$ , we get $f'(0) = -f'(4) = f'(4)$ , so $f'(0) = f'(4) = 0$ .

By Rolle's theorem, for $n = 1,2,3,4,5,6$ , there exists a $x_n \in (n-1,n)$ such that $f''(x_n) = 0$ .

Thus, the function $g(x) = f'(x)f''(x)$ has zeros at $x = 0,x_1,1,x_2,2,x_3,3,x_4,4,x_5,5,x_6,6$ .

By using Rolle's theorem again, $g'(x) = f'(x)f'''(x)+f''(x)^2$ has at least $12$ zeros in $[0,6]$ .

Therefore, the minimum number of zeros of $f'(x)f'''(x)+f''(x)^2$ on $[0,6]$ is $12$ .