Answer:
(f + g)(x)≥3 for all values of x
Step-by-step explanation:
Given the expressions f(x) = |x| + 9 and g(x) = –6, sine f(x) contains the absolute value of a variable x, this absolute value can be negative and positive. Therefore f(x) can be expressed in two forms as shown;
f(x) = x+9 and f(x) = -x+9
If f(x) = x+ 9 and g(x) = -6
(f + g)(x) = f(x)+ g(x)
(f + g)(x) = x+9+(-6)
(f + g)(x) = x+9-6
(f + g)(x) = x+3
Similarly, if f(x) = -x+ 9 and g(x) = -6
(f + g)(x) = f(x)+ g(x)
(f + g)(x) = -x+9+(-6)
(f + g)(x) = -x+9-6
(f + g)(x) = -x+3
(f + g)(x) = 3-x
In both expresson, we have bith x to be positive and negative, hence we can write the value of resulting x as an absolute value as shown;
(f + g)(x) = |x|+3
This shows that (f + g)(x)≥3 for all values of x
