SOLUTION: https://gyazo.com/1062ba896258a824814c5eedd6c4c4f9

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Question 1187088: https://gyazo.com/1062ba896258a824814c5eedd6c4c4f9
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The requirements for f(x) are...

(1) has a local minimum
(2) has two negative intervals
(3) is an even function

From (3), we know the exponents on all variables must be even.

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

%28x%5E2-a%5E2%29+=+%28x-a%29%28x%2Ba%29

That function has vertical asymptotes at x=-a and x=a.

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.

So we want a function of the form

%28-b%29%2F%28x%5E2-a%5E2%29+=+%28-b%29%2F%28%28x-a%29%28x%2Ba%29%29

where a and b are positive constants.

That function has the following characteristics:

(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

So all the requirements are met.

After playing with the numbers to find a function with a nice graph, I came up with the function

f%28x%29=%28-64%29%2F%28x%5E2-16%29

Here is a graph, showing all the required characteristics....

graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C-64%2F%28x%5E2-16%29%29