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xmas
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Find the inflection points at x=C and x=D with C less than or equal to D?
Consider the function f(x) = x^(2)e^(9x).
I just don't know how to determine what the inflection points are after I find the second derivative.
f(x) has two inflection points at x = C and x = D with C less than or equal to D
What is C
What is D.
I
 2 years ago
 2 years ago
xmas Group Title
Find the inflection points at x=C and x=D with C less than or equal to D? Consider the function f(x) = x^(2)e^(9x). I just don't know how to determine what the inflection points are after I find the second derivative. f(x) has two inflection points at x = C and x = D with C less than or equal to D What is C What is D. I
 2 years ago
 2 years ago

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SomeBloke Group TitleBest ResponseYou've already chosen the best response.0
At an inflection point, the second derivative is zero. For example, consider cos(x) at x=π/2, since that's an inflection point. \[\frac{ d }{ dx }\cos x =\sin x\]\[\frac{ d^{2} }{ dx ^{2} }\cos x =\cos x\]\[\cos \frac{ \pi }{ 2 } =0\]At that point, the first derivative, sin x, is at a maximum, so the tangent at that point is a horizontal line. Since the first derivative is either at a maximum or a minimum at a point of inflection (on the original function), the second derivative will be zero at that same point.
 2 years ago

Sujay Group TitleBest ResponseYou've already chosen the best response.0
Test for concavity first: if the 2nd derivative of f is more than 0, then it is concave up, if it is less than 0, concave down. If concavity changes at a critical value, then it is an inflection point. The critical values in this sense would be whenever the second derivative is equal to 0. Test the intervals that apply between the critical values. If concavity changes, then there you have it, an inflection point. If the second derivative = 0 on that interval, than the second derivative test fails, and you must try the first derivative test.
 2 years ago
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