what point of f(x)= e^(1-x^2) is concave down? I think it is 1/2 but I'm not sure
To determine where the function f(x) = e^(1-x^2) is concave down, we need to find the second derivative and analyze its sign. Here's how you can do it:
1. Start by finding the first derivative of f(x) with respect to x.
f'(x) = d/dx (e^(1-x^2))
Using the chain rule, the derivative of e^(1-x^2) can be written as:
f'(x) = e^(1-x^2) * (-2x)
2. Now, differentiate f'(x) to find the second derivative.
f''(x) = d/dx (f'(x))
f''(x) = d/dx (e^(1-x^2) * (-2x))
Applying the product rule, the derivative can be written as:
f''(x) = -2e^(1-x^2) + 2x(2x * e^(1-x^2))
3. Simplify the expression for f''(x):
f''(x) = -2e^(1-x^2) + 4x^2 * e^(1-x^2)
Now, we need to determine the values of x for which f''(x) is negative, indicating concavity down.
To find these points, set f''(x) less than zero and solve for x:
-2e^(1-x^2) + 4x^2 * e^(1-x^2) < 0
Divide throughout by e^(1-x^2) (which is always positive):
-2 + 4x^2 < 0
Rearrange the inequality:
4x^2 - 2 < 0
Factor the expression:
2(2x^2 - 1) < 0
Now solve for x:
2x^2 - 1 < 0
Add 1 to both sides:
2x^2 < 1
Divide by 2:
x^2 < 1/2
Now, take the square root of both sides:
|x| < sqrt(1/2)
Considering both positive and negative values of x, we conclude that the function f(x) = e^(1 - x^2) is concave down for all values of x where |x| < sqrt(1/2) or -sqrt(1/2) < x < sqrt(1/2).
Therefore, your initial guess that x = 1/2 is incorrect. The function is actually concave down for values of x between -sqrt(1/2) and sqrt(1/2).