Question

If the set of real roots of the function f(x) is {-2,1,3}, find the set of real roots of f(x^2) and f(x^3-6)

Answer 1

\[f(x)=(x+2)(x-1)(x-3)\\
f(x^2)=(x^2+2)(x^2-1)(x^2-3)\]

Answer 2

\[x^2+2=0\] has no real solution
\[x^2-1=0\iff x =\pm1\\
x^2-3=0\iff x=\pm\sqrt3\]

Answer 3

f(-2) = 0; f(1) = 0; f(3) = 0 (given)

Take f(-2) = 0
f(x^3-6) will be zero when x^3 - 6 = -2 or x^3 = 4 or \(\large x = \sqrt[3]{4}\)

Take f(1) = 0
f(x^3-6) will be zero when x^3 - 6 = 1 or x^3 = 7 or \(\large x = \sqrt[3]{7}\)

Try the case of f(3) = 0
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