o.k., with the purpose to do this, you may take the 2d by-product and discover its roots(what makes it=0) hence, i'm getting the 2d by-product to be... f''=(8e^(-4x))(2x-a million) this might equivalent 0 while x=a million/2 and while x=infinity So there are 2 feasible effects... i)x<a million/2 Take any variety decrease than a million/2 and plug it in and spot what sign you get. enable's do 0 considering the fact that its elementary You get 8(-a million)=-8 So it incredibly is CONCAVE DOWN for any variety decrease than a million/2 ii) x>a million/2 Plug in a million for simplicity You get 8e^(-4)(2-a million)=8e^(-4). So it incredibly is CONCAVE UP for and x>a million/2 You answer is any x decrease than a million/2 There you circulate!!!

## Answers & Comments

## Verified answer

Mechanically, you find the 2nd derivative of y and find the values of x that make y'' < 0.

y'' = 16x e^(-4x) - 8 e^(-4x) = 8 e^(-4x)(2x-1)

Notice that 8 e^(-2x) can never be less than 0 for any x.

So the question is when is 2x - 1 < 0 ==> x < 1/2

Hence the graph is concave down for x in the interval

(- \infty, 1/2)

o.k., with the purpose to do this, you may take the 2d by-product and discover its roots(what makes it=0) hence, i'm getting the 2d by-product to be... f''=(8e^(-4x))(2x-a million) this might equivalent 0 while x=a million/2 and while x=infinity So there are 2 feasible effects... i)x<a million/2 Take any variety decrease than a million/2 and plug it in and spot what sign you get. enable's do 0 considering the fact that its elementary You get 8(-a million)=-8 So it incredibly is CONCAVE DOWN for any variety decrease than a million/2 ii) x>a million/2 Plug in a million for simplicity You get 8e^(-4)(2-a million)=8e^(-4). So it incredibly is CONCAVE UP for and x>a million/2 You answer is any x decrease than a million/2 There you circulate!!!

it looks like it is concave down on the entire domain. if anything, its from(-infinity,.25) then undefined on the rest