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

1 Answer
Jan 1, 2018

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]}