Question

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

Answer 1

The function is convex for `x in (-oo,0) uu(0, +oo)`

Explanation:

`"Reminder"`

`(u/v)'=(u'v-uv')/v^2`

`(uv)'=u'v+uv'`

Calculate the first and second derivatives

`f(x)=e^x/(5x^2)+1`

`f'(x)=(e^x*(5x^2)-e^x*(10x))/(25x^4)=((x-2)e^x)/(5x^3)`

`(x-2)'=1`

`((x-2)e^x)'=e^x+(x-2)e^x=e^x(x-1)`

`f''(x)=(e^x(x-1)(5x^3)-(x-2)e^x(15x^2))/(25x^6)`

`=(e^x(x^2-x-3x+6))/(5x^4)`

`=(e^x(x^2-4x+6))/(5x^4)`

The inflection points are when `f''(x)=0`

The discriminant of the quadratic equation `x^2-4x+6=0`

is

`Delta=b^2-4ac=(-4)^2-4*(1)*(6)=16-24=-8`

As the discriminant is negative, the second derivative

`f''(x)>0`, `AA x in RR -{0}`

The variation chart is as follows

`color(white)(aaaa)` `"Interval"` `color(white)(aaaa)` `(-oo,0)` `color(white)(aaaa)` `(0, +oo)`

`color(white)(aaaa)` `"Sign f''(x)"` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaaaaa)` `"f(x)"` `color(white)(aaaaaaaa)` `uu` `color(white)(aaaaaaaa)` `uu`

graph{e^x/(5x^2)+1 [-9.71, 10.29, -1.96, 8.04]}