f(x)=e^(-6x^2) Where is it concave up? On what interval?
Find the first and second derivatives
f’ = (-12x)(e^(-6x^2)) = -12xe^(-6x^2)
f’’= (-12x)( 12xe^(-6x^2))+( e^(-6x^2))(-12)
=144x^2e^(-6x^2) - 12e^(-6x^2)
=12e^(-6x^2)(12x^2-12)
Set the first derivative equal to zero to get critical point
-12xe^(-6x^2) = 0
e^(-6x^2) = 0 (e can never =0)
-12x=0
X=0
Plug that into second derivative result is -12 which means at the point
X=0 the curve is concave down
Now set the second derivative equal to zero to get the inflection points
12e^(-6x^2)(12x^2-12)=0
12xe^(-6x^2) = 0 (e can never =0)
12x^2-12=0
Add 12 to both sides
12x^2=12
Divide by 12
X^2=1
Square root both sides and x=-1 and +1
So the function is concave down -1<x<1
The function is concave up from where x is negative infinity to -1 and from 1 to positive infinity
x<-1 and x>1
