Answer:
(a) Domain: All real numbers except -2
(b) [tex]f(x) > -4[/tex]
Step-by-step explanation:
Given
See attachment for f(x)
Solving (a): The domain
From the attached image, we have:
[tex]f(x) = -x - 2, x <-2[/tex]
[tex]f(x) = -x^2, -2 < x < 0[/tex]
[tex]f(x) = x, x \ge 0[/tex]
To get the domain, we consider the inequalities attached to each of the piece-wise function
[tex]x < -2[/tex]
This implies that the values of x is less than -2 i.e.
[tex]x = \{.......,-4,-3\}[/tex]
[tex]-2< x<0[/tex]
This implies that x is greater than -2 and less than 0 i.e.
[tex]x = \{-1\}[/tex]
[tex]x \ge 0[/tex]
This implies that x is greater than or equal to 0 i.e.
[tex]x =\{0,1,2,3,....\}[/tex]
If these values of x are merged, we have:
[tex]x = \{.......,-4,-3,-1,0,1,2,3.....\}[/tex]
[tex]x = \{-\infty,..,-4,-3,-1,0,1,2,3,...,\infty\}[/tex]
-2 is not included in the above values of x.
Solving (b): The range
From the attached image, we have:
[tex]f(x) = -x - 2, x <-2[/tex]
[tex]f(x) = -x^2, -2 < x < 0[/tex]
[tex]f(x) = x, x \ge 0[/tex]
Substitute the greatest value of x in each piece-wise function.
[tex]f(x) = -x - 2, x <-2[/tex]
[tex]x=-2[/tex]
So;
[tex]f(-2) = -2-2 = -4[/tex]
[tex]f(x) = -x^2, -2 < x < 0[/tex]
[tex]x = -2[/tex]
So:
[tex]f(-2) = -2^2 = -4[/tex]
[tex]f(x) = x, x \ge 0[/tex]
[tex]x = \infty[/tex]
So;
[tex]f(\infty) = \infty[/tex]
We have:
[tex]f(-2) =-4[/tex]
[tex]f(-2) =-4[/tex]
[tex]f(\infty) = \infty[/tex]
The smallest value of f(x) is -4
Hence, the range is:
[tex]f(x) > -4[/tex]
Answer:
Domain is
B)
(-infinity, -2) U (-2, infinity)
Range is (-4, infinity)
Step-by-step explanation:
go it correct
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