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]}