# Increasing and Decreasing Intervals

1. Jan 14, 2005

### courtrigrad

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 $$f'(x) = 1+6e^(-x^2)xln(e) [/tex[. Then I set this equal to 0 But I get [tex] {e = e, x = RootOf(.(1+6*exp(-ln(e)_Z^2)_Zln(e) = 0, _Z))}$$

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.

Thanks a lot

2. Jan 14, 2005

### dextercioby

It's this the function??
$$y=y(x)=x-3e^{-x^{2}}$$

If so,review its differentiation.

Daniel.

3. Jan 15, 2005

### HallsofIvy

Staff Emeritus
Assuming that $y(x)= x- 3e^{-x^2}$ then $y'= 1+ 6xe^{-x^2}$.
(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.)

4. Jan 15, 2005

### courtrigrad

thanks a lot Halls and dexter

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