Finding zero of function problem

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Homework Statement



the function f(x)=tan(3^x) has one zero in the interval [0,1.4]. The derivative at this point is


The Attempt at a Solution



f'(x)= sec^2(3^x)*x3^(x-1) and i really can't seem to get a zero from that because i really don't know where to start and if I am misinterpreting the problem please tell me that. Thanks
 
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Wait, what is the initial question?
Are you trying to find roots of:

y=\tan{(3^x})
http://img7.imageshack.us/img7/7748/tan3x.jpg
 
Last edited by a moderator:
If that's the case, your derivative is incorrect.

f'(x) = 3^x \log(3) \sec^2({3^x})
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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