What are the domain and range of f (x) = (one-fifth) Superscript x?
2 answers:
Answer:
The domain of the function is all real numbers and the range is all positive real numbers
Step-by-step explanation:
We have the following function and we want to find the domain and the range.
The function we have is an example of an exponential function with b > 0 and b ≠ 1. This types of functions in general have the following properties:
- It is always greater than 0, and never crosses the x-axis
- Its domain is the set of real numbers
- Its Range is the Positive Real Numbers
The domain of a function is the specific set of values that the independent variable in a function can take on.
When determining domain it is more convenient to determine where the function would not exist.
This function has no undefined points nor domain constraints. Therefore the domain is .
The range is the resulting values that the dependent variable can have as x varies throughout the domain. Therefore the range is .
We can check our results with the graph of the function.
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Answer:
The answer is below
Step-by-step explanation:
A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.
a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.
a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?
Answer:
Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)
The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:
(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:
(x - 3)² + (y - 4)² = 3²
(x - 3)² + (y - 4)² = 9
The diagram is drawn using geogebra.
b) The house is 5 miles north. It can be represented by y = 5 line.
To find the distance east of the house we have to substitute y = 5 and solve for x, hence:
(x - 3)² + (y - 4)² = 9
(x - 3)² + (5 - 4)² = 9
(x - 3)² + 1 = 9
(x - 3)² = 8
(x - 3) = √8
x - 3 = ±2.83
x = 3 ± 2.83
x = 5.83 or 1.83
Since it is to the east to the cell tower, hence x = 5.83.
Therefore the house is located 5.83 miles to the east of the cell tower
