If , for how many real numbers does ?

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The correct answer is E.

To determine how many real numbers satisfy is to determine how many solutions the equation has, which in turn is to determine how many solutions the equation has. It is not difficult to see, using the Rational Roots Theorem, that has no factor of the form for any whole or rational number value of , but there are real number values of such that is a factor. This problem must be solved using a non-algebraic method.

One way to determine how many numbers satisfy is to examine the graph of the function . Use a graphing calculator to graph for on a suitably large interval to see all intersections of the graph with the line , and then count the number of points of intersection.

There are at least four such points: A first point with -coordinate between and , a second point with -coordinate between and , a third point with -coordinate between and and a fourth point with -coordinate between and . The fact that is a polynomial of degree means that there can be at most four such points. Therefore, there are four values of for which $f(k)=2$.

Alternatively, one can examine a table of values of the function and then identify intervals for which the values of at the endpoints have different signs and apply the Intermediate Value Theorem to each of those intervals. Create a table of values for for whole number values of between and , inclusive, and count the number of intervals of length for which the value of is greater than for one of the endpoints and less than for the other endpoint.

There are four such intervals: , , and . By the Intermediate Value Theorem, for each of these four intervals there is a value in that interval such that . This shows that there are at least four such values of and the fact that is a polynomial of degree means that there can be at most four such values of Therefore, there are four values of for which .