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if $f(x)$ be non constant thrice differentiable function defined on $(-\infty,\infty)$ such that

$f(x) = f(6-x)$ and $f'(0) = f'(2) = f'(5) =0.$ Determine the minimum number of

zeros of $g(x) = (f''(x))^2+f'(x)f'''(x)$ in $[0,6]$

Attempt: from $f(x) = f(6-x),$ substitute $x\rightarrow x+3$

so $f(3+x) = f(3-x)$ (function $f(x)$ is symmetrical about $x=3$ line)

wan,t be able to go further, could some help me

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    Note that $g(x)$ is the derivative of $f'(x)f''(x)$. – 2017-02-14

1 Answers 1

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Let $h(x)=f'(x)f''(x)$.

Now, we can use symmetry to find that $f'(x)$ has a minimum of $7$ solutions in $[0,6]$.

Then, we have that $f''(x)$ has a minimum of $6$ solutions in $[0, 6]$ from Rolle's Theorem.

So, we have that there are at least $13$ different solutions to $h(x)=0$.

By Rolle's Theorem, we have that $g(x)=h'(x)$ has a minimum of $12$ different solutions.

EXAMPLE OF SUCH A FUNCTION

Try $$\cos \pi x$$

We get that $g(x)$ has $12$ different zeroes in $[0,6]$.

Thus, we are done. The answer is $12$.