Posted by Kay on Monday, September 16, 2013 at 1:38pm.
How many distinct real numbers x are true for f(x)=0. Explain.
F(x)=x^2+(1/(500x^21000))

Math  Steve, Monday, September 16, 2013 at 2:13pm
x^2 + 1/(500(x^22)) = 0
If f(x) were x^2, then there would be a double root near 0. But, the graph is shifted down just a smidgen, so there will now be two roots near zero. Also, since there are asymptotes at ±√2, the graph will cross the xaxis near ±√2, making two more roots.
Or, we can adjust things a bit to work with a straight polynomial.
x^2 (x^22) + 1/500 = 0
If f(x) were x^2 (x^22) then there would be a double root at x=0 and roots at x = ±√2.
But our f(x) is translated up by 1/500, meaning we still have two roots near ±√2, but also now two roots near x=0. So, 4 distinct roots in all.
Answer This Question
Related Questions
 PreCalculus  A piece of equipment has cost function C(x)=50x^2 + 1000 and its ...
 Math  Assume that f is a continuous function from the real numbers to the real ...
 Algebra 2  Sarah has a job that pays $45,500 per year with a guaranteed pay ...
 Math  Which set of best describes the numbers used on a scale for a standard ...
 Math  For how many ordered triples of positive real numbers (a,b,c), do the ...
 Math  For how many ordered triples of positive real numbers (a,b,c), do the ...
 math: real numbers  Determine whether the following statement is true or fase. ...
 Algebra II  I have a question The range of y = 1/(x5) is All real Numbers ...
 math  What is the minimum value of N that will make this statement true: If we ...
 Math Check Pleas Thank You  1. Which of the following numbers is an example of ...
More Related Questions