Question 1187088
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The requirements for f(x) are...<br>
(1) has a local minimum
(2) has two negative intervals
(3) is an even function<br>
From (3), we know the exponents on all variables must be even.<br>
To get a local minimum, we want a rational function that is always positive between two vertical asymptotes.  To get two vertical asymptotes in an even rational function, we can have a denominator of the form<br>
{{{(x^2-a^2) = (x-a)(x+a)}}}<br>
That function has vertical asymptotes at x=-a and x=a.<br>
Given asymptotes at x=-a and x=a, to get a local minimum we want the function to be positive everywhere between x=-a and x=a; in particular, we want it to be positive at x=0.  If we put just a positive constant in the numerator, then the function value is negative at x=0, so we can put a negative constant in the numerator.<br>
So we want a function of the form<br>
{{{(-b)/(x^2-a^2) = (-b)/((x-a)(x+a))}}}<br>
where a and b are positive constants.<br>
That function has the following characteristics:<br>
(1) It is even
(2) It has vertical asymptotes at x=-a and x=a
(3) Its value is positive at x=0
(4) With vertical asymptotes at x=-a and x=a and a positive value at x=0, it has a local minimum
(5) When x is less than -a or greater than a, the function value is negative, so its graph has two negative intervals<br>
So all the requirements are met.<br>
After playing with the numbers to find a function with a nice graph, I came up with the function<br>
{{{f(x)=(-64)/(x^2-16)}}}<br>
Here is a graph, showing all the required characteristics....<br>
{{{graph(400,400,-10,10,-10,10,-64/(x^2-16))}}}<br>