Questions 1-4 : Find the domain and range of the function.
2. g(x) = -3(x + 1) 2 + 6
5. A gull landing on the guardrail causes a pebble to fall off the edge. The speed of the pebble as it falls to the ground is a function is v(d) = √(2gd) where d is the distance, in meters, the pebble has fallen, v(d) is the speed of the pebble, in meters per second (m/s) and g is the acceleration due to gravity—about 9.8 meters per second squared (m/s 2 ). Determine the domain and range of v(d), the pebble’s speed.
6. Vitaly and Sherry have 24 m of fencing to enclose a rectangular garden at the back of their house.
a) Express the area of the garden as a function of its width.
b) Determine the domain and range of the area function.
This is a linear function, so x and y can be any value.
This is a quadratic equation in vertex form. The function has a maximum value at the vertex (-1, 6), x can be any value.
We cannot take the square root of a negative number, so (2 - x) must be positive or zero.
√(2 - x ) means the positive square root, so y is never negative.
The given function is a rational function.
To find the domain of a rational function, we have to find the value of x that makes the denominator zero.
In 1/(x - 2), if we substitute x = 2, the denominator becomes zero and it is undefined.
So, p(x) is defined for all real values of x except x = 2.
Domain of p(x) = R -
To find range of the rational function above, find the inverse of p(x).
Interchange the variables.
Solve for y in terms of x.
Find the domain of p -1 (x).
In (2x + 1)/x , if we substitute x = 0, the denominator becomes zero and it is undefined.
So, p -1 (x) is defined for all real values of x except x = 0.
Domain of p -1 (x) = R -
Range of p(x) = Domain of p -1 (x)
When d = 0 the pebble begins to fall, and d = 346 when it lands.
So, the domain is 0 ≤ d ≤ 346.
The pebble starts with speed 0 m/s.
v(0) = v(d) = √(2 ⋅ 9.8 ⋅ 0 ) = 0
When the pebble lands, d = 346.
v(346) = √(2 ⋅ 9.8 ⋅ 346 )
and the range is
They need fencing on only three sides of the garden because the house forms the last side.
Let the width of the garden be x m.
Then the length is (24 - 2x).
Let A(x) be the area of the garden.
= -2[x 2 - 2(x)(6) + 6 2 - 6 2 ]
A(x) = -2(x - 6) 2 - 72
The smallest the width can approach is 0 m. The largest the width can approach is 12 m.
A(x) = -2(x - 6) 2 - 72 is a quadratic equation in vertex form and it has a maximum value at the vertex (6, 72).
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