Question

Given f(x)= x square+ 6x + 12 show that f(x) ≥3 for al values of x
the answer plz faaassst

Answer 1

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)