The function is
#f(x)=(e^x)/(x^2+1)#
Calculate the first and second derivatives
The first derivative is the derivative of a quotient
#(u/v)=(u'v-uv')/v^2#
#u=e^x#, #=>#, #u'=e^x#
#v=5x^2+1#, #=>#, #v'=10x#
#f'(x)=(e^x(5x^2+1)-e^x(10x))/(5x^2+1)^2#
#=(e^x(5x^2-10x+1))/(5x^2+1)^2#
The second derivative is the derivative of a quotient
#(u/v)=(u'v-uv')/v^2#
#u=e^x(5x^2-10x+1)#, #=>#, #u'=e^x(5x^2-10x+1)+10e^x(x-1)#
#v=(5x^2+1)^2#, #=>#, #v'=2*5x*(5x^2+1)=10x(5x^2+1)#
Therefore,
#f''(x)=((e^x(5x^2-10x+1)+10e^x(x-1))*(5x^2+1)^2-(e^x(5x^2-10x+1))(10x(5x^2+1)))/(5x^2+1)^4#
#=(e^x((5x^2-10x+1+10x-10)(5x^2+1)-10x(5x^2-10x+1)))/(5x^2+1)^3#
#=(e^x(25x^4-100x^3+160x^2-20x-9))/(5x^2+1)^3#
The points of inflections are when
#f''(x)=0#
#=>#, #(e^x(25x^4-100x^3+160x^2-20x-9))/(5x^2+1)^3=0#
#=>#, #25x^4-100x^3+160x^2-20x-9=0#
Solving #25x^4-100x^3+160x^2-20x-9=0# graphically
graph{25x^4-100x^3+160x^2-20x-9 [-3.465, 3.464, -1.73, 1.734]}
The points of inflection are #(-0.175, 0.712)# and #(0.355, 0.86)#
Let's build a variation chart to determine the concavities
#color(white)(aa)##"Interval"##color(white)(aa)##(-oo,-0.175)##color(white)(a)##(-0.175,0.355)##color(white)(a)##(0.355,+oo)#
#color(white)(aa)##"Sign f''(x)"##color(white)(aaaaaa)##+##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaaaaa)##+#
#color(white)(aaaa)##"f(x)"##color(white)(aaaaaaaaa)##uu##color(white)(aaaaaaaaaaaaaa)##nn##color(white)(aaaaaaaaa)##uu#
The function is convex in the intervals #x in (-oo,-0.175) uu(0.355,+oo)# and concave in the interval #x in (-0.175,0.355)#
graph{e^x/(5x^2+1) [-3.465, 3.464, -1.73, 1.734]}
