SOLUTION: Consider the following function.
f (x)  =  (1/sqrt{4π})*e^(−x^2/18)
(note that 4π is under the square root in the denominator)
Find the la
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-> SOLUTION: Consider the following function.
f (x)  =  (1/sqrt{4π})*e^(−x^2/18)
(note that 4π is under the square root in the denominator)
Find the la
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Question 1107577: Consider the following function.
f (x) = (1/sqrt{4π})*e^(−x^2/18)
(note that 4π is under the square root in the denominator)
Find the largest interval(s) on which f is concave up. Answer by greenestamps(13200) (Show Source):
(1) The function value is always positive
(2) The function has its maximum value when x=0
(3) The function is an even function, symmetric with respect to the y-axis.
(4) The limit of the function is 0 when x goes to positive or negative infinity
From those observations, we know the graph will be concave down between -a and a for some positive value a and concave up everywhere else. So there will be two largest intervals where the function is concave up -- from negative infinity to a, and from a to infinity.
We now only need to find the value of a; for simplicity we will work with positive values of x.
The graph changes from concave down to concave up when the second derivative is zero -- that is, when the slope changes from decreasing to increasing.
f(x) =
f'(x) = [chain rule, and derivative of e^f(x)]
f''(x) = [product rule, chain rule, and derivative of e^f(x)]
The second derivative is zero when
The value of a we are looking for is 3; the two largest intervals on which the graph is concave up are from negative infinity to -3 and from 3 to infinity.
