Question 1107577
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Do some analysis first....<br>
(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<br>
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.<br>
We now only need to find the value of a; for simplicity we will work with positive values of x.<br>
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.<br>
f(x) = {{{(1/sqrt(4*pi))e^((-1/18)*x^2)}}}<br>
f'(x) = {{{(1/sqrt(4*pi))(e^((-1/18)*x^2))((-1/9)x) = (-x/(9*sqrt(4*pi)))(e^((-1/18)*x^2))((1/9)x)}}}  [chain rule, and derivative of e^f(x)]<br>
f''(x) = {{{(-1/(9*sqrt(4*pi)))(e^((-1/18)*x^2))+(-x/(9*sqrt(4*pi)))(e^((-1/18)*x^2))((-1/9)x) = (e^((-1/18)*x^2)/(9*sqrt(4*pi)))*(-1+x^2/9)}}}[product rule, chain rule, and derivative of e^f(x)]<br>
The second derivative is zero when
{{{-1+x^2/9 = 0}}}
{{{x^2/9 = 1}}}
{{{x^2 = 9}}}
{{{x = 3}}}<br>
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.