QueenCarrot
contestada

Hi, another math problem :
A cubic function is definined for x∈R as :
f(x)= x³ +x(1-k²) + k , where k is constant.
a) Show that -k is root of f.
b) Find, in term of k, the other two roots of f.
c) Find the set of values of k for which f has exactly one real root.

Thanks a lot

Respuesta :

MathisLif3
f=x^3+x(1-k^2)+k
f(-k)=-k^3-k(1-k^2)+k
=-k^3-k+k^3+k
=0
in terms of k, we can divide by x+k since its a root
we get x^2-kx+1=0, find the two roots
x=(k-sqrt(k^2-4))/2 and (k+sqrt(k^2-4))/2

f has one root in case k is contained in (-2,2)

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