For what values of x is #f(x)= e^x/(5x^2 +1# concave or convex?

Question

Ezekiel Alexander · Accepted Answer

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)#

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]}

Axel Alvarez · Answer

undefined The function f(x) = e^x / (5x^2 + 1) is convex for all values of x.