William C. answered • 09/27/23
Experienced Tutor Specializing in Chemistry, Math, and Physics
1. f(x) = (x² – 81)/(x + 9)
We can't have a zero denominator, so the function is undefined at x = –9.
All other values of x are allowed.
Domain: (–∞,–9) ∪ (–9,∞)
An (almost) equivalent form of the function is given by
(x² – 81)/(x + 9) = [(x + 9)(x – 9)]/(x + 9) = x – 9
So f(x) = x – 9 for all x ≠ –9
This means that –9 – 9 = –18 cannot be in the range of f(x)
Range: (–∞,–18) ∪ (–18,∞)
Sketch of f(x):
A sketch of f(x) will look like the line y = x – 9, but with one difference. The point (–9,–18) is missing.
(see desmos.com/calculator/hmvnytfeel)
2. f(x) = √(9-x²)
We can't have a negative number under the radical, so the function is undefined at x > 3 and x < –3.
All other values of x are allowed.
Domain: [–3,3]
f(x) = √(9-x²) ≥ 0
Range: [0,∞)
Sketch of f(x):
If we take y = √(9-x²) and square both side we get y² = 9 – x² which means that x² + y² = 9
The relation x² + y² = 9 describes a circle that is centered at (0,0) and a radius of 3.
Since the range of f(x) = √(9-x²) has a range of [0,∞), its sketch will be a semicircle, in the upper half of the coordinate plan, that is centered at (0,0) and a radius of 3.
(see desmos.com/calculator/hmvnytfeel)
3. f(x) = {x² – 1 if x < 3
{x + 5 if x ≥ 3
This is not what was written, but I'm assuming this is what was meant.
There are no restrictions on allowed values of x, so the domain is all real numbers
Domain: (–∞,∞)
Looking at the two parts of the piecewise function we see
For f(x) = x² – 1 if x < 3
since x² ≥ 0, this means f(x) ≥ –1
For x + 5 if x ≥ 3
f(x) ≥ 8
This means that the range of f(x) is all numbers greater than or equal to –1
Range: [–1,∞)
Sketch of f(x):
For –∞ < x < 3 we have a parabola shaped like y = x² with its vertex at (0,–1) instead of (0,0)
For 3≤ < x < ∞ we have a ray with a slope of 1 starting at the point (3,8) and slanting upward to the right.
(see desmos.com/calculator/hmvnytfeel)
