Answer:
The range of values of [tex](f + g)(x) \ge 3[/tex] is for all values of x [tex]-\infty[/tex] to [tex]+\infty[/tex]
Step-by-step explanation:
Given
[tex]f(x) = |x| + 9[/tex]
[tex]g(x) = -6[/tex]
Required
Describe the range of [tex](f+g)(x)[/tex]
First, calculate [tex](f+g)(x)[/tex]
[tex](f+g)(x) = f(x) + g(x)[/tex]
Substitute values for f(x) and g(x)
[tex](f+g)(x) = |x| + 9 -6[/tex]
Evaluate Like Terms
[tex](f+g)(x) = |x| + 3[/tex]
The above expression shows that [tex](f + g)(x) \ge 3[/tex]
Because [tex]|x| \ge 0[/tex]
Hence, the range of values of [tex](f + g)(x) \ge 3[/tex] is for all values of x [tex]-\infty[/tex] to [tex]+\infty[/tex]
