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August 17th, 2016, 03:37 PM   #1
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Critical Points

Hello

I'm calculating critical points, so it's f'(x)=0

f'(x)= (4/3)x- tanx
then
(4/3)x-tanx=0

Wolfram alpha states that one of the answers is +-(0.844)

I tried rewriting it as:

(4/3)xcosx-sinx=0 ; by rewriting tanx as sinx/cosx

but I couldn't think of a numerical solution.

Is there any way to solve this using algebra? or any trig trick that I may be missing.

Thanks
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August 17th, 2016, 04:17 PM   #2
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$f'(x) = \dfrac{4}{3}x - \tan{x} = 0$

$f'(x)$ is an odd function, so zeros will be symmetrical to the y-axis.

$x = 0$ is a solution ... any others will require a numerical method
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