Given f(x)= x square+ 6x + 12 show that f(x) ≥3 for al values of x
the answer plz faaassst
1 answer:
Step-by-step explanation:
f(x) = x² + 6x + 12
= (x² + 6x + 9) + 3
= (x + 3)² + 3.
Since a² >= 0 for all real values of a,
(x + 3)² >= 0 for all real values of x,
(x + 3)² + 3 >= 3 for all real values of x.
Hence f(x) >= 3. (Shown)
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