Real Zeros of a Quartic Family with a Parameter
Let f(x) = x⁴ − 4x³ + 4x² + a, where a is a real parameter. The graph of y = f(x) is being studied for a modelling problem in which the number and location of real zeros determine the qualitative behaviour of a system. You may use calculus freely, but you should justify the classification by referring to the signs of f at its critical points rather than by drawing a graph. The parameter a can be any real number; the case a = −1 is of particular interest because the system is then at a threshold between two qualitative regimes. Note that x = 1 is suspected to be a repeated zero in that threshold case. You are asked to identify the exact parameter values at which the number of distinct real zeros changes.
1. Classify the number of distinct real zeros of f as a function of a, giving the critical values of a at which the count changes.
Read the worked answer
Differentiate: f′(x) = 4x³ − 12x² + 8x = 4x(x − 1)(x − 2). Critical points are x = 0, 1, 2.
f(0) = a, f(1) = a + 1, f(2) = a.
The sign of f′ is negative on (−∞,0), positive on (0,1), negative on (1,2), and positive on (2,∞). Hence x = 0 and x = 2 are local minima with value a, and x = 1 is a local maximum with value a + 1.
The number of distinct real zeros changes when a local extremum touches the x-axis, i.e. at a = 0 (minima touch) and a = −1 (maximum touches).
Classification: • a > 0: all three critical values are positive; f(x) > 0 for all x, so 0 real zeros. • a = 0: f(x) = x²(x − 2)²; zeros x = 0 and x = 2, each double; 2 distinct real zeros. • −1 < a < 0: minima are negative and the maximum is positive; the curve crosses once on each of the four monotonic intervals; 4 distinct real zeros. • a = −1: the maximum equals 0 and the minima are negative; 3 distinct real zeros (one double at x = 1 plus two simple). • a < −1: the maximum is negative and the minima are more negative; the curve crosses once on each of the two outer intervals; 2 distinct real zeros.
Thus the counts change at a = −1 and a = 0.
2. For a = −1, factor f completely over the reals and state the multiplicity of each zero. Confirm that this is the threshold at which the number of distinct real zeros changes.
Read the worked answer
At a = −1, f(x) = x⁴ − 4x³ + 4x² − 1 = [x(x − 2)]² − 1.
Let u = x(x − 2) = x² − 2x. Then f = (u − 1)(u + 1).
So f = (x² − 2x − 1)(x² − 2x + 1).
The second factor is (x − 1)². The first factor has roots x = [2 ± √(4 + 4)]/2 = 1 ± √2.
Therefore f(x) = (x − 1 − √2)(x − 1 + √2)(x − 1)².
Zeros: x = 1 (double), x = 1 + √2, and x = 1 − √2. There are three distinct real zeros.
This is the threshold a = −1 because the local maximum at x = 1 has just reached zero. For a slightly greater than −1 the maximum is positive and two extra crossings appear near x = 1, giving four distinct zeros; for a slightly smaller than −1 the maximum is negative and those two crossings disappear, leaving two distinct zeros.