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Calculus and Beyond Homework Help
How can I determine the potential extrema on a function with a discontinuity?
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[QUOTE="fishspawned, post: 5996845, member: 382901"] [h2]Homework Statement [/h2] i am looking for the potential extrema on the function y = |sinx(x)| + (1/x) between [-2pi, 2pi] [h2]Homework Equations[/h2] extrema will be likely located at f'(x) = 0 or u/d [h2]The Attempt at a Solution[/h2] first it is noted that there is a discontinuity at x=0 then determining f'(x) by using |sin(x)| as sqrt(sin[SUP]2[/SUP]x) instead we get f'(x) = sin(x)cos(x)/|sinx(x)| - x[SUP]-2[/SUP] first the easy one : we can see that f'(x) = u/d where sin(x) = 0 [looking only at the denominator here] that will be at x= -2pi, -pi, pi, and 2pi My problem is the second part f'(x) = 0 when the whole equation = 0 0 = sin(x)cos(x)/|sinx(x)| - x[SUP]-2[/SUP] therefore sin(x)cos(x)/|sinx(x)| = x[SUP]-2[/SUP] using |sin(x)| as sqrt(sin[SUP]2[/SUP]x) instead we get sin(x)cos(x)/(sqrt(sin[SUP]2[/SUP]x)) = x[SUP]-2[/SUP] a little algebra and then we get x[SUP]4[/SUP] = sec[SUP]2[/SUP](x) and now... i am stuck. is there a way to solve this particular equation? [/QUOTE]
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How can I determine the potential extrema on a function with a discontinuity?
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