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Suppose that the function $f(x)$ satisfies $f(0)=0$, $f'(0)=0$ and $f''(0)>0$.

(a) Show that there exists $d>0$ such that $\frac{f'(x)-f'(0)}{x-0}>0$ for every non-zero $x$.

(b) Use Rolle's Theorem to show that $f(x)$ is not equal to $0$ for all non-zero x in $(-d,d)$.


I'm okay with part (a).

My problem with part (b) is that using Rolle's Theorem here assumes continuity on some closed intervals, say $[a,0]$ and $[0,b]$, and differentiability on the corresponding open intervals $(a,0)$ and $(0,b)$. Obviously $f(x)$ is continuous at $x=0$, but how does that guarantee the conditions needed for the application of Rolle's theorem? I have completed part (b) by assuming the required conditions but would like to know if/why I am justified in doing so.

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The existence of $f''(0)$ requires that there be an open interval $(-\delta, \delta)$ where $f'$ be defined. (i.e., since $\lim_{t \to 0} \frac{f'(t)}{t}$ exists, this follows from the requirements which the limit definition imposes.) –  user1296727 Jan 4 '13 at 7:36
    
Also: there appears to be a typographical error in (a), or an omission. –  user1296727 Jan 4 '13 at 7:41
    
Ok, yes, I think I understand. The existence of the second derivative at x=0 requires the definition of the first derivative on (-d, d), except for at x=0 (and we already know f '(0) exists). The definition of f '(x) on (-d,d) in turn imply continuity of f on (-d,d). Is that right? Thanks. –  jim Jan 4 '13 at 7:55
    
Yes, $\exists f''(0) \implies \exists \delta > 0: f'$ is defined on $[-\delta, \delta] \implies f$ is continuous on $[-\delta, \delta]$. (To obtain a closed interval $[-\delta, \delta]$, just restrict the open interval.) –  user1296727 Jan 4 '13 at 8:15
    
(You understand quantifiers, right? When I write $\exists$, I mean "there exists". Also: by putting your math in dollar signs, i.e. (dollarsign) f(x) (dollarsign), you get back $f(x)$. (dollarsign) \frac{a}{b} (dollarsign) will get you $\frac{a}{b}$, etc. You can right-click on other people's math and see how they wrote it, too.) –  user1296727 Jan 4 '13 at 8:22
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