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I came across the above problem. I see that $G'(x)=f'(x)f(\sqrt{tan(f(x))})$. Now since $f$ is a differentiable even function, $f(-x)=f(x)$ and so $-f'(-x)=f'(x)$ and thus $f'(0)=0$ and hence we can conclude $G'(0)=0.$ Am I going in the right direction?

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2  
If $f(0)<0$, $G$ does not defiened as real function. –  tetori Jan 17 '13 at 4:10
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Differentiation not quite right, is $f'(x)\sqrt{\tan(f(x))}$. –  André Nicolas Jan 17 '13 at 4:11
    
Thanks a lot sir for pointing out the mistake.I have corrected my post. –  learner Jan 17 '13 at 4:33
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could you post the source of your questions like this? –  Une Femme Douce Jan 17 '13 at 4:45

2 Answers 2

up vote 1 down vote accepted

The answer is not always $B$ either. Following Babak Sorouh's idea, let $f(x) = \pi/2 - x^2$. Then

$$ G'(x) = -2 x \sqrt{\tan\!\left(\pi/2-x^2\right)} $$

where it exists and

$$ \lim_{x \to 0} G'(x) = -2. $$

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The example can be modified to have $f(x)$ positive everywhere since all that matters here is the behavior in a neighborhood of $x=0$. –  Antonio Vargas Jan 17 '13 at 7:02
    
Thanks a lot sir for the detailed clarification. I have got it. –  learner Jan 17 '13 at 10:02
    
Great example Thanks for sharing me that. +1 –  Babak S. Jan 18 '13 at 3:31

Firstly, note that according to Leibniz rule, $G'(x)=f'(x)\sqrt{\tan(f(x))}$. Secondly, buy taking $f(x)=x^{2n}$ as a family of even functions with real valued, we have: $$G'(0)=0$$ so A and C cannot be true for a general case under problem's assumptions.

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This doesn't say whether B or D is true. For that, learner had a good idea, but as tetori points out it may depend on whether we add an assumption of positivity of $f$. –  Jonas Meyer Jan 17 '13 at 4:25
    
@JonasMeyer: Yes, exactly. Since we are facing a multiple choices here so I think we don't need to prove it theoretically. And, as tetori point and that is absolutely true, it seems D looks right. I am thinking of an even function in which $\sqrt{\tan(f(0))}$ is undefined. –  Babak S. Jan 17 '13 at 5:35
    
Nice lead, and hits home, I think! +1 –  amWhy Feb 15 '13 at 3:02
    
@amWhy: Thanks for considering this small hint. dear amWhy, my credit for accessing the web has been expired, and I will have the new one 2 days later I think. That's why I have not been here. –  Babak S. Feb 15 '13 at 8:11

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