When $\frac{dy}{dx}=e^{-2x}(1-2x)$, stationary point will be when $x=\frac{1}{2}$? How come in my book, theres an inflection point when $x=1$? I thought inflection point is also a stationary point? So it should be gotten by setting $\frac{dy}{dx}=0$, which got me only $x=\frac{1}{2}$?
Stationary/Inflection Point(s) when $\frac{dy}{dx}=e^{-2x}(1-2x)$
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ordinary-differential-equations
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0[classification](http://en.wikipedia.org/wiki/Stationary_point#Classification) – 2011-11-18
1 Answers
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Stationary points are where f'(x)=0. So for your example, $x=1/2$ is the stationary point.
Inflection points occur when the graph of $f$ switches concavity. This occurs when f'' changes sign. Candidates for inflection points are found by finding points where f''(x)=0 or not exist.
The tangent line at an inflection point need not have 0 slope. Look at $y=\tan x$ at $x=0$, or your example.
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0I should point out that inflection points occur at points $x=c$ where $f'$ has a local extreme value. So, the candidates for inflection points of $f$ are the stationary points of $f'$ (these are the points where $(f')'=f''$ is zero) or points where $(f')'=f''$ does not exist. – 2011-11-18