Increasing and Decreasing Intervals

AI Thread Summary
To analyze the function y = x - 3e^-x^2 for increasing and decreasing intervals, the first step is to find its derivative, which is y' = 1 + 6xe^-x^2. The function is increasing where the derivative is positive and decreasing where it is negative, determined by setting the derivative to zero. For concavity, the second derivative is used, indicating concave upward when positive and concave downward when negative, with inflection points occurring where the second derivative changes sign. It’s essential to confirm the conditions for inflection points, as the second derivative being zero is a necessary but not sufficient condition.
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Hello all

If you are given the function y = x - 3e^-x^2 and you want to find the intervals where the function is increasing and decreasing, concavity, inflection points and any local extreme values, would I first find the derivative?

My work

If f(x) = x - 3e^-x^2 then f&#039;(x) = 1+6e^(-x^2)xln(e) [/tex[. Then I set this equal to 0 But I get {e = e, x = RootOf(`.`(1+6*exp(-ln(e)_Z^2)_Zln(e) = 0, _Z))}<br /> <br /> How would you determine concavity and any local extrema? I know that to get inflection points you take the second derivative and set it equal to 0.<br /> <br /> Thanks a lot
 
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It's this the function??
y=y(x)=x-3e^{-x^{2}}

If so,review its differentiation.

Daniel.
 
Assuming that y(x)= x- 3e^{-x^2} then y&#039;= 1+ 6xe^{-x^2}.
(You really have it right: ln(e)= 1).

A function is increasing where its derivative is positive, decreasing where its derivative is negative. Of course, those intervals are separated by points where the derivative is equal to 0 so the first thing you would do is determine where the derivative is 0: where [/itex]1+ 6xe^{-x^2}= 0[/itex].

A function is "concave upward" if its slope is getting greater: i.e. where its derivative is increasing so its second derivative is positive. A function is concave downward where its second derivative is negative and has an inflection point where its second derivative changes sign. Note that the second derivative must be 0 at an inflection point but that is not sufficient! (y= x4 does not have an inflection point at x= 0.)
 
thanks a lot Halls and dexter
 
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