Answer:
a. Domain: {x | x is all real numbers} Range: {y | y ≥ 6}
Step-by-step explanation:
This is a modular function. All values inside brackets will result in absolute values. In other words, positive numbers.
And we can solve this problem algebraically and graphically.
f(x) = |x-3|+6
There are no restrictions and this function is continuous since we can proceed and take it out the brackets as we'll proceed.
I) y =|x-3|+6
Assuming the symmetrical values for x and -3, i.e. -x and +3
y=-(x-3)+6
y= -x+3+6
y=-x+9
We have a linear equation with no restriction in the Set of Real Numbers. We can choose any value for x, we'll have another for y.
II) Assuming the same values we did in the brackets.
y= x-3 +6
y=x+3
We have another linear equation with no restriction. We can choose any value for x, we'll have another for y.
Finally, we can affirm that the
Domain is a number that belongs to this Interval: (-∞,∞) inside the Set of Real Numbers or
D={x∈R} even or just like the way it's been written {x|x is all Real Numbers}
Range
The range of function depends on the value set for x since it's the result of y.
But we can say where it starts. And in case of restrictions where it may not continue.
In this case, the range of the function starts at:
y=|x-3|+6
As the range starts at 6, precisely the point (0,6) no matter which value you pick for x. As the function continues with no restriction and goes on and on we can say the Range belongs to this interval [6, ∞) or {y ∈ R /y≥6}
