作业帮12.Let f(x) = x^2 + 6x + c for all real numbers x,where c is
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12.Let f(x) = x^2 + 6x + c for all real numbers x,where c is some real number.For what values of c does f(f(x)) have exactly 3 distinct real roots?
Suppose f has only one distinct root r1.Then,if x1 is a root of f(f(x)),it must be the case that f(x1) = r1.As a result,f(f(x)) would have at most two roots,thus not satisfying the problem condition.Hence f has two distinct roots.Let them be r1 ≠ r2.Since f(f(x)) has just three distinct roots,either f(x) = r1 or f(x) = r2 has one distinct root.Assume without loss of generality that r1 has one distinct root.Then f(x) = x^2 +6x+c = r1 has one root,so
that x^2 + 6x + c - r1 is a square polynomial.Therefore,c - r1 = 9,so that r1 = c - 9.So c - 9 is a root of f.So (c - 9)^2 + 6(c -9) + c = 0,yielding c2 - 11c + 27 = 0,or (c - 11/2 )^2 = 13/2 .This results to c = (11±√13)/2 .
that x^2 + 6x + c - r1 is a square polynomial.Therefore,c - r1 = 9,so that r1 = c - 9.So c - 9 is a root of f.So (c - 9)^2 + 6(c -9) + c = 0,yielding c2 - 11c + 27 = 0,or (c - 11/2 )^2 = 13/2 .This results to c = (11±√13)/2 .
12.Let f(x) = x^2 + 6x + c for all real numbers x,where c is
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