# CalculusGiven the function y=t*e^[-t^2+1], find the max and mins.

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### 1 Answer

`y = te^(-t^2+1)` **Differentiating wrt t.**

`y' = 1(e^(-t^2+1))+t(-2t)(e^(-t^2+1))`

`y' = (1-2t^2)e^(-t^2+1)`

**For extreme points (maxima, minima and inflection points),**

`y' = 0`

`(1-2t^2) = 0`

`t = +-1/sqrt(2)`

So we have two points `t=1/sqrt(2)` and `-1/sqrt(2)` . To determine whether they are maxima or minima we need to find the sign of `y''` .

`y'' = (-4t)(e^(-t^2+1))+(1-2t^2)(-2t)(e^(-t^2+1))`

`y'' = (4t^3-6t)(e^(-t^2+1))`

`y'' = 2t(2t^2-3)(e^(-t^2+1))`

At `t = 1/sqrt(2)` , **y'' = negative**, therefore at `t = 1/sqrt(2)` , y has a maximum.

**The maximum value is** `y = 1/sqrt(2)sqrte`

At `t = -1/sqrt(2)` , **y'' is positive**, therefore at `t = 1/sqrt(2)` , y has a minimum.

**The minimum value is** `y = -1/sqrt(2)sqrte`