Function f(x)= e^-x^2
a) Inflection Values
b) Intervals on which f(x) is concave downward
c) Intervals on which f(x) is concave upward.
If you look at the graph of this function you will see a standard "bell curve"
f'(x) = -2x(e^(-x^2))
f''(x) = -2x(-2x)(e^(-x^2)) - 2(e^(-x^2))
= 4x^2(e^(-x^2)) - 2(e^(-x^2))
= 2e^(-x^2)(2x^2 - 1)
(e^(-x^2)) = 0
1/(e^(x^2))= 0 -----> no solution
or
2x^2 - 1 = 0
x = ± 1/√2
sub that back into f''(x) to find the values at the points of inflection
b) f(x) is concave down when f''(x) < 0
f''(x) = 2(e^(-x^2))(2x^2-1)
clearly (e^(-x^2)) or 1/(e^(x^2)) is always positive, so all we need to look at is the 2x^2 -1 part
concave up for -1/√2 < x < 1/√2
c) mmmmhhh?
To find the inflection points of the function f(x) = e^(-x^2), we need to determine the values of x where the concavity changes. Inflection points occur when the second derivative of a function changes sign.
a) Inflection values:
1. Find the first derivative of f(x): f'(x) = -2xe^(-x^2).
2. Find the second derivative of f(x): f''(x) = (4x^2 - 2)e^(-x^2).
3. Set f''(x) = 0 and solve for x: (4x^2 - 2)e^(-x^2) = 0.
This equation has two solutions, x = -sqrt(2)/2 and x = sqrt(2)/2.
4. Plug these x-values back into f(x) to obtain the corresponding y-values:
f(-sqrt(2)/2) ≈ 0.6065 and f(sqrt(2)/2) ≈ 0.6065.
Therefore, the inflection points are (x = -sqrt(2)/2, y = 0.6065) and (x = sqrt(2)/2, y = 0.6065).
b) Intervals on which f(x) is concave downward:
For a function to be concave downward, the second derivative must be negative.
1. We know from above that f''(x) = (4x^2 - 2)e^(-x^2).
2. To find the intervals where the second derivative is negative, set f''(x) < 0 and solve for x:
(4x^2 - 2)e^(-x^2) < 0.
3. The solution to this inequality is -sqrt(2)/2 < x < sqrt(2)/2.
Therefore, the function f(x) = e^(-x^2) is concave downward in the interval (-sqrt(2)/2, sqrt(2)/2).
c) Intervals on which f(x) is concave upward:
For a function to be concave upward, the second derivative must be positive.
1. We know from above that f''(x) = (4x^2 - 2)e^(-x^2).
2. To find the intervals where the second derivative is positive, set f''(x) > 0 and solve for x:
(4x^2 - 2)e^(-x^2) > 0.
3. The solutions to this inequality are x < -sqrt(2)/2 and x > sqrt(2)/2.
Therefore, the function f(x) = e^(-x^2) is concave upward in the intervals (-∞, -sqrt(2)/2) and (sqrt(2)/2, ∞).