Let $f$ be a non negative function and has a finite third derivative in the open interval $(0,1)$, then $f'''(c) = 0$ for some $c \in (0,1)$ if :
(a) $f(x) =0$ for atleast two values of $x \in (0,1)$
(b) $f(x) = 0$ for atleast one value of $x \in (0,1)$
(c) $f'(x) = 0$ for atleast two values of $x \in (0,1)$
(d) None of these
Using a counterexample $f(x) = (x – 1/2)^4$ we can say options (a) and (c) are clearly false, But I am not able
to prove/disprove option (b)
Any hints ??
Best Answer
(a) is true: If $0<a<b<1$ and $f(a)=0=f(b),$ then $f$ has minimums at $a,b.$ This implies $f'(a)=f'(b)=0.$ But note also $f$ has a local maximum at some $c\in (a,b),$ where $f'(c)=0.$ Thus $f'=0$ at three distinct points. The MVT then implies $f''$ at two distinct points. Apply MVT again to see $f'''=0$ somewhere.
(b) is false: Let $f(x)=(x-1/2)^2+(x-1/2)^3.$ Then $f(1/2)=0,$ $f>0$ on $(0,1),$ and $f'''\equiv 6.$
Hint for (c): Let $0<a<b<1.$ Think of $f'(x)=(x-a)(x-b).$