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)uucolor(white)(aaaaaaaaaaaaaa)nncolor(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]}
