Answer :
The function is f(x) = e^(-x^2).
The first step is to take the first derivative, remember to use the chain rule
d/dx[f(g(x))]=f '(g(x)) x g '(x).
f '(x) = e^(-x^2) x (-2x)
f '(x) = -2xe^(-x^2)
We have now calculated the first derivative f '(x) = -2xe^(-x^2).
Now, calculate the 2nd derivative, remember the product rule
d/dx[f(x)g(x)]=f '(x)g(x)+g '(x)f(x)
f "(x) = -2e^(-x^2)-2x(-2xe^(-x^2))
f "(x) = -2e^(-x^2)+4x^2(e^(-x^2))
f "(x) = (e^(-x^2))(4x^2-2)
We have now calculate the second derivative f "(x)=(e^(-x^2))(4x^2-2).
Next, you are going to want to set our second derivative equal to zero to calculate possible inflection points.
(e^(-x^2))(4x^2-2) = 0
e^(-x^2)=0 4x^2-2 = 0
-x^2=ln(0) 4x^2=2
ln(0) is undefined, but you can still
continue to solve 4x^2=2
x^2=1/2
x=1/2(-sqrt(2)), x=1/2(sqrt(2))
So, (-1/2(sqrt(2)),1/e(sqrt(e))) and (1/2(sqrt(2)),1/e(sqrt(e))) are the possible inflection points.
By this point what you want to do is plug in numbers in these three intervals
(-infinity,-1/2(sqrt(2))), (-1/2(sqrt(2)),1/2(sqrt(2))) and (1/2(sqrt(2)),infinity), I would choose -1, 0 and 1. But remember to plug these numbers into the second derivative of our original function.
f "(-1) = 2/e which is positive, this means our function f(x) = e^(-x^2) is concave up before we reach our first possible inflection point.
f "(0) = -2 which is negative, this means our function is concave down before we reach second possible inflection point.
f "(1) = 2/e which is positive. Since both f "(-1) and f "(1) are positive, this means we have verified that (-1/2(sqrt(2)),1/e(sqrt(e))) and (1/2(sqrt(2)),1/e(sqrt(e))) are actual inflection points. Between these two inflection points our function is concave down. Therefore the function f(x) = e^(-x^2) is concave down on the interval (-1/2(sqrt(2)),1/2sqrt(2))) using interval notation. Or using inequality notation -1/2(sqrt(2))<x<1/2(sqrt(2))
The first step is to take the first derivative, remember to use the chain rule
d/dx[f(g(x))]=f '(g(x)) x g '(x).
f '(x) = e^(-x^2) x (-2x)
f '(x) = -2xe^(-x^2)
We have now calculated the first derivative f '(x) = -2xe^(-x^2).
Now, calculate the 2nd derivative, remember the product rule
d/dx[f(x)g(x)]=f '(x)g(x)+g '(x)f(x)
f "(x) = -2e^(-x^2)-2x(-2xe^(-x^2))
f "(x) = -2e^(-x^2)+4x^2(e^(-x^2))
f "(x) = (e^(-x^2))(4x^2-2)
We have now calculate the second derivative f "(x)=(e^(-x^2))(4x^2-2).
Next, you are going to want to set our second derivative equal to zero to calculate possible inflection points.
(e^(-x^2))(4x^2-2) = 0
e^(-x^2)=0 4x^2-2 = 0
-x^2=ln(0) 4x^2=2
ln(0) is undefined, but you can still
continue to solve 4x^2=2
x^2=1/2
x=1/2(-sqrt(2)), x=1/2(sqrt(2))
So, (-1/2(sqrt(2)),1/e(sqrt(e))) and (1/2(sqrt(2)),1/e(sqrt(e))) are the possible inflection points.
By this point what you want to do is plug in numbers in these three intervals
(-infinity,-1/2(sqrt(2))), (-1/2(sqrt(2)),1/2(sqrt(2))) and (1/2(sqrt(2)),infinity), I would choose -1, 0 and 1. But remember to plug these numbers into the second derivative of our original function.
f "(-1) = 2/e which is positive, this means our function f(x) = e^(-x^2) is concave up before we reach our first possible inflection point.
f "(0) = -2 which is negative, this means our function is concave down before we reach second possible inflection point.
f "(1) = 2/e which is positive. Since both f "(-1) and f "(1) are positive, this means we have verified that (-1/2(sqrt(2)),1/e(sqrt(e))) and (1/2(sqrt(2)),1/e(sqrt(e))) are actual inflection points. Between these two inflection points our function is concave down. Therefore the function f(x) = e^(-x^2) is concave down on the interval (-1/2(sqrt(2)),1/2sqrt(2))) using interval notation. Or using inequality notation -1/2(sqrt(2))<x<1/2(sqrt(2))