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OpenStudy (anonymous):
Find the inflection point of: f(x) = e^{cosx} on [0,π], correct to 6 decimal places I know: f'(x) = e^(cosx)*(-sinx) f''(x) = e^(cosx)*(-cosx) + e^(cosx)*(-sinx)(-sinx) = e^(cosx)*(-cosx) + e^(cosx)*(sinx)^2 Then, I factor out e^(cosx) and let f''(x) = 0 e^(cosx)[(sinx)^2 - cosx] = 0 So I would solve for (sinx)^2 - cosx = 0 and find the x value for the inflection point?
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OpenStudy (anonymous):
assuming that your work is correct, and only going to the last line, solve \[\sin^2(x)-\cos(x)=0\] by replacing \(\sin^2(x)\) by \(1-\cos^2(x)\) and solving \[1-\cos^2(x)-\cos(x)=0\]
OpenStudy (anonymous):
For the x value I got x = 0.904557. I printed the function and its first 2 deriv.
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