Question

For what values of x is `f(x) = e^-(1/x^2)` concave or convex?

Answer 1

See explanation

Explanation:

`f(x)=e^(−(1/x^2))=e^(−x^(-2))`

`f'(x)=e^(−(1/x^2))*2x^(-3)`

`f''(x)=e^(−(1/x^2))(2x^(-3))^2+e^(−(1/x^2))*(-6x^(-4))`

`f''(x)=e^(−(1/x^2))[4x^(-6)-6x^(-4)]`

`f''(x)=e^(−(1/x^2))*2x^(-4)(2x^(-2)-3)`

`f''(x)=(2)/(e^(1/x^2)x^(4))*((2-3x^2)/x^(2))`

Everything is always positive except `2-3x^2`

`2-3x^2=0`

`3x^2-2=0`

`3(x^2-2/3)=0`

`3(x-sqrt(2/3))(x+sqrt(2/3))=0`

`3(x-sqrt6/3)(x+sqrt6/3)=0`

`x=+-sqrt6/3~~+-0.816`

`x in (-oo,-sqrt6/3) hArr f''(x)<0quad` f is concave `nn`

`x in (-sqrt6/3,sqrt6/3) hArr f''(x)>0quad` f is convex `uu`

`x in (sqrt6/3,oo) hArr f''(x)<0quad` f is concave `nn`

graph{e^(-1/(x^2)) [-2.294, 2.57, -0.83, 1.603]}