Luis, Diego, and Cecil are going fishing.
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Two main facts are needed here:
1. The logarithm , regardless of the base of the logarithm, exists for
.
2. The square root exists for
.
(in both cases we're assuming real-valued functions only)
By (2) we know that exists if
, or
.
By (1), we know that exists if
, or
. But as long as the square root exists, it will always be positive, so this condition will always be met.
Ultimately, then, we only require , so the function has domain
.
To determine the range, we need to know that, in their respective domains, and
increase monotonically without bound. We also know that
at minimum, at which point the square root term vanishes, so the least value the function takes on is
. Then its range would be
.
Greetings!
To start this problem, let's first assign a variable for the missing, consecutive odd numbers. Since they are consecutive and odd, we add two.
<u>Proof:</u> <em>3-1=2, 5-3=2</em>
The first, consecutive, odd number:
The second, consecutive, odd number:
The third, consecutive, odd number:
The fourth, consecutive, odd number:
The sum of the values are equal to 3 times the sum of the first two numbers, of which this is equal to 35 less than the fourth number. Let's create an equation to simplify this:
Complete the operations inside the parenthesis:
Distribute the parenthesis (utilizing the distributive property)
Simplify both sides:
Add -6 and -x to both sides of the equation:
Divide both sides of the equation by 5:
If is equal to -7:
The four numbers are:
I hope this helps!
-Benjamin