1.) Find the domain and range for:

f(x)= 4x-20/ the square root of 36-x^2

2.) Solve the equation:

log6 (x-11) + log6 (x-6) = 2

3.) Solve the inequality:

x^2 - 2x - 3 is greater than 0

1. remember that the denominator √(36 - x^2) cannot be zero or negative,

so shouldn't -6 < x < +6 ??

2. log6 [(x-11)(x-6)] = 2
(x-11)(x-6) = 6^2
x^2 - 17x + 66 = 36
x^2 - 17x + 30 = 0
(x-15)(x-2) = 0
x = 15 or x = 2

A lot of students would stop here and think they have the right answer, but remember that we can only take logs of positive numbers, so looking at our original we can see clearly that x > 11

so x = 15 is the only answer.

3. x^2 - 2x - 3 > 0

(x-3)(x+1) > 0
"critical" values are x = 3 and x = -1

try a number < -1, say x = -5
statement: (-8)(-4) > 0 ? YES
try a number between -1 and 3, say x = 0
statement: (-3)(1) > 0 False!
try a number > 3, say x = 5
statement: (2)(6)>0 YES

so x < -1 OR x > 3, x any real number.

1.) To find the domain and range of the function f(x) = (4x - 20) / sqrt(36 - x^2), we need to consider two things:

Domain: The denominator, sqrt(36 - x^2), cannot be equal to zero since division by zero is undefined. Therefore, we should find the values of x that make the denominator equal to zero. Solving for sqrt(36 - x^2) = 0, we get 36 - x^2 = 0, which implies x^2 = 36. Taking the square root of both sides, we have x = ±6.

Thus, the domain is all real numbers except x = ±6. In interval notation, the domain can be expressed as (-∞, -6) ∪ (-6, 6) ∪ (6, ∞).

Range: To find the range, we need to determine the possible values of f(x) as x varies. As x approaches -6 or 6, the denominator approaches zero, and f(x) approaches positive or negative infinity, respectively. Since the numerator is 4x - 20, it continues to increase or decrease without bound as x approaches infinity or negative infinity.

Therefore, the range of the function f(x) is (-∞, ∞).

2.) To solve the equation log6 (x - 11) + log6 (x - 6) = 2, we can use the properties of logarithms to simplify it first.

Using the product rule of logarithms, we can combine the two logarithms on the left side as a single logarithm:
log6[(x - 11)(x - 6)] = 2.

To eliminate the logarithm, we can rewrite the equation in exponential form:
6^2 = (x - 11)(x - 6).

Simplifying, we have:
36 = (x - 11)(x - 6).

Expanding the right side, we get:
36 = x^2 - 17x + 66.

Rearranging the equation, we have a quadratic equation in standard form:
x^2 - 17x + 30 = 0.

Factoring the quadratic equation, we find:
(x - 2)(x - 15) = 0.

Setting each factor equal to zero, we have two possible values for x:
x - 2 = 0 or x - 15 = 0.

Solving for x, we get:
x = 2 or x = 15.

Therefore, the equation log6 (x - 11) + log6 (x - 6) = 2 is satisfied when x = 2 or x = 15.

3.) To solve the inequality x^2 - 2x - 3 > 0, we can first find the roots of the corresponding equation x^2 - 2x - 3 = 0.

Using factoring or the quadratic formula, we can find that the roots are x = -1 and x = 3.

Next, we can plot these roots on a number line:

<-----(-1)------(3)------->

Now, we want to determine the sign of the quadratic expression x^2 - 2x - 3 in each interval. We can choose a number in each of the intervals to test:

For x < -1, we can choose x = -2. Substituting this into the inequality:

(-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5, which is positive.

For -1 < x < 3, we can choose x = 0. Substituting this into the inequality:

(0)^2 - 2(0) - 3 = 0 - 0 - 3 = -3, which is negative.

For x > 3, we can choose x = 4. Substituting this into the inequality:

(4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5, which is positive.

Based on these test results, we can see that the inequality x^2 - 2x - 3 > 0 is satisfied when x < -1 or x > 3.

Therefore, the solution to the inequality is x < -1 or x > 3.

1.) To find the domain and range of a function, we need to consider which values are allowed for the independent variable (x) and the corresponding values for the dependent variable (y).

a) Domain:
In this case, we need to examine the expression inside the square root, 36-x^2. For the square root to be defined, the value inside the square root must be greater than or equal to 0. So, we have the inequality 36-x^2 ≥ 0.

To solve this inequality, we need to factorize it:
(6+x)(6-x) ≥ 0

The critical points occur when (6+x) = 0 or (6-x) = 0, which gives us x = -6 and x = 6.

We can now create a sign chart to determine the values of x that satisfy the inequality:
Test point: x = -7 (any value less than -6)
(6+x)(6-x) = (-)(-) = (+) [greater than 0]

Test point: x = 0 (between -6 and 6)
(6+x)(6-x) = (+)(-) = (-) [less than 0]

Test point: x = 7 (any value greater than 6)
(6+x)(6-x) = (+)(+) = (+) [greater than 0]

Based on the sign chart, we see that the expression (36-x^2) is positive (greater than 0) for x < -6 and x > 6. However, we need to consider the square root, and it is important to note that the square root of 0 is 0. Therefore, the square root of (36-x^2) is also defined at x = -6 and x = 6.

Since the original function is f(x) = 4x-20/√(36-x^2), the domain is all real numbers except x = -6 and x = 6.

b) Range:
The range refers to the set of possible values for the dependent variable (y). In this case, the dependent variable is given by y = f(x) = 4x-20/√(36-x^2).

As x approaches -6 (from either side), the value inside the square root approaches 0, making the denominator arbitrarily small, and the function tends towards positive or negative infinity.

Therefore, the range of this function is (-∞, ∞), which means it can take any real value.

2.) To solve the equation log6(x-11) + log6(x-6) = 2:

We can use the properties of logarithms to simplify the equation. The sum of logarithms with the same base is equal to the logarithm of the product.

log6[(x-11)(x-6)] = 2

Now, we can rewrite the equation in exponential form using the definition of logarithms. In exponential form, the base of the logarithm is the base of the exponential expression and the value of the logarithm is the exponent.

6^2 = (x-11)(x-6)

36 = x^2 - 17x + 66

Rearranging the equation:
x^2 - 17x + 30 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula.

Factoring:
(x-2)(x-15) = 0

From this, we can conclude that x = 2 or x = 15.

Therefore, the solutions to the equation are x = 2 and x = 15.

3.) To solve the inequality x^2 - 2x - 3 > 0:

We need to identify the values of x that make the expression greater than 0.

First, factorize the quadratic expression:

(x-3)(x+1) > 0

Now, set each factor to zero and solve for x:

x - 3 = 0 --> x = 3
x + 1 = 0 --> x = -1

Plot the points x = -1 and x = 3 on a number line:

-∞ (-1) 3 +∞

Now, we need to test different intervals on the number line to determine the inequality's sign.

Pick a test point from each interval and substitute it into the inequality. For example:
For x < -1, choose x = -2:
(-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0

For -1 < x < 3, choose x = 0:
(0 - 3)(0 + 1) = (-3)(1) = -3 < 0

For x > 3, choose x = 4:
(4 - 3)(4 + 1) = (1)(5) = 5 > 0

Based on the test points, we can see that the inequality is greater than 0 for x < -1 or x > 3.

The solution to the inequality x^2 - 2x - 3 > 0 is x < -1 or x > 3.