PLEASE HELP!!
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Answer:
First, we know that the function IxI works as follows:
IxI = x if x ≥ 0
IxI = -x if x < 0
So IxI is always positive.
then:
a) f(x) = Ix - 2I
this is:
f(x) = (x - 2) if (x - 2) ≥ 0
-(x - 2) if (x - 2) < 0
Then the graph of this function will be two lines (where the separation of the lines depends on the vale of x)
For the case of b we have
g(x) = -IxI + 1
then we have:
g(x) = -x + 1 if x ≥ 0
g(x) = -(-x) + 1 if x < 0
To graph the functions, you need to graph these lines for the given values of x. You also can see the graphs below. Where the red graph is the one for the point 4a, and the blue graph is the one for point 4b.
Now, let's find the domain and range:
4a) f(x) = Ix - 2I
The domain is the set of the possible values of x that we can input in that function. For this case, we do not have any value of x that generates a problem (like a zero in a denominator, for example) then the domain is the set of all real numbers: x ∈ R
The range is the set of the possible values of y.
f(x) = Ix - 2I is always equal or greater than zero (as you can also see in the graph) then the range will be:
R: y ∈ R, such that y ≥ 0.
4b) g(x) = -IxI + 1
Similar as the prior case, here the domain is the set of all real numbers:
D: x ∈ R
For the range, now we have:
y = -IxI + 1
if IxI is always equal or larger than zero, then -IxI is always equal or smaller than zero. If we add a + 1 to that, then the function will be always equal or smaller than 1.
Then the range is:
R: y ∈ R, such that y ≤ 1
