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Question 21 of 32

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SAT Subject Test

If f left parenthesis x right parenthesis equals x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4, for how many real numbers k does f left parenthesis k right parenthesis equals 2 invisible times?

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None

Correct Answer: 
No

One

Correct Answer: 
No

Two

Correct Answer: 
No

Three

Correct Answer: 
No

Four

Correct Answer: 
Yes

The correct answer is E.

To determine how many real numbers k satisfy f left parenthesis k right parenthesis equals 2 is to determine how many solutions the equation x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 equals 2 has, which in turn is to determine how many solutions the equation x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 2 equals 0 has. It is not difficult to see, using the Rational Roots Theorem, that x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 2 has no factor of the form x minus k for any whole or rational number value of k, but there are real number values of k such that x minus k is a factor. This problem must be solved using a non-algebraic method.

One way to determine how many numbers k satisfy f left parenthesis k right parenthesis equals 2 is to examine the graph of the function f left parenthesis x right parenthesis equals x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 invisible times. Use a graphing calculator to graph y equals x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 for x on a suitably large interval to see all intersections of the graph with the line y equals 2, and then count the number of points of intersection.

The figure presents a line and a curve in the x y plane with the origin labeled O. The numbers negative 5 and 5 are indicated on the x-axis. The numbers negative 50 and 50 are indicated on the y-axis. The line is horizontal and lies above the x-axis. It crosses the y-axis at the point with coordinates 0 comma 2. The curve begins high above the x-axis and to the left of the y-axis between x values of negative 3 and negative 2. It curves downward and to the right, crosses the line and then the x-axis between x values of negative 2 and negative 1, and continues to a minimum slightly below the x-axis between x values of negative 2 and negative 1. The curve then moves upward and to the right, crosses the x-axis and then the line between x values of negative 1 and 0, and reaches a maximum on the y-axis slightly above the line. The curve moves downward again, crosses the line and then the x-axis between x values of 0 and 1, and continues to a minimum far below the x-axis, between x values of 3 and 4. Finally, it curves upward, crosses the x-axis and then the line between x values of 4 and 5, and ends high above the x-axis and to the right of the x value of 5.

There are at least four such points: A first point with x-coordinate between negative 2 and negative 1, a second point with x-coordinate between negative 1 and 0, a third point with x-coordinate between 0 and 1 and a fourth point with x-coordinate between 4 and 5. The fact that x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4  is a polynomial of degree 4 means that there can be at most four such points. Therefore, there are four values of k for which f ( k ) = 2 .

Alternatively, one can examine a table of values of the function f left parenthesis x right parenthesis equals x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 and then identify intervals for which the values of f left parenthesis x right parenthesis minus 2 at the endpoints have different signs and apply the Intermediate Value Theorem to each of those intervals. Create a table of values for y equals x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 for whole number values of x betweennegative 5 and 5, inclusive, and count the number of intervals of length 1 for which the value of f is greater than 2 for one of the endpoints and less than 2 for the other endpoint.

The figure presents a 2-column table with 11 rows of data. The heading for column 1 is “x.” The heading for column 2 is “f of x.” The data are as follows. Row 1: x equals negative 5; f of x equals 779 Row 2: x equals negative 4; f of x equals 308 Row 3: x equals negative 3; f of x equals 85 Row 4: x equals negative 2; f of x equals 8 Row 5: x equals negative 1; f of x equals negative 1 Row 6: x equals 0; f of x equals 4 Row 7: x equals 1; f of x equals negative 7 Row 8: x equals 2; f of x equals negative 40 Row 9: x equals 3; f of x equals negative 77 Row 10: x equals 4; f of x equals negative 76 Row 11: x equals 5; f of x equals 29

There are four such intervals: left square bracket negative 2 comma negative 1 right square bracket, left square bracket negative 1 comma 0 right square bracket, left square bracket 0 comma 1 right square bracket and left square bracket 4 comma 5 right square bracket. By the Intermediate Value Theorem, for each of these four intervals there is a value k in that interval such that f left parenthesis k right parenthesis equals 2. This shows that there are at least four such values of k comma and the fact that x to the power of 4 end exponent minus 3 x to the power of 3 end exponent minus 9 x to the power of 2 end exponent plus 4 is a polynomial of degree 4 means that there can be at most four such values of k. Therefore, there are four values of k for which f left parenthesis k right parenthesis equals 2.

Question Difficulty: 
medium