Answer :
To determine which equation correctly represents the cost [tex]\( f(x) \)[/tex] of shipping a package that weighs [tex]\( x \)[/tex] pounds, let's break down the problem and analyze each of the given possible equations.
1. Base Cost and Additional Cost:
- The base cost is [tex]\( \$6 \)[/tex] for packages weighing up to 1 pound.
- For each additional pound or portion of a pound, the cost increases by [tex]\( \$2 \)[/tex].
2. Analyzing the Equations:
We'll evaluate the costs for different values of [tex]\( x \)[/tex], including fractions around 1 pound, to determine which equation fits the given cost structure.
### Equation Analysis
1. Equation 1: [tex]\( f(x) = 6 + 2 \lceil x - 1 \rceil \)[/tex], where [tex]\( x > 0 \)[/tex]
- For [tex]\( x = 0.5 \)[/tex] pounds:
[tex]\[ f(0.5) = 6 + 2 \lceil 0.5 - 1 \rceil = 6 + 2 \lceil -0.5 \rceil = 6 + 2(-1) = 6 - 2 = 4 \][/tex]
- This satisfies the given cost structure since 0.5 would be rounded to 1, resulting in an additional cost.
2. Equation 2: [tex]\( f(x) = 6 + 2 \lfloor x - 1 \rfloor \)[/tex], where [tex]\( x > 0 \)[/tex]
- For [tex]\( x = 0.5 \)[/tex] pounds:
[tex]\[ f(0.5) = 6 + 2 \lfloor 0.5 - 1 \rfloor = 6 + 2 \lfloor -0.5 \rfloor = 6 + 2(-1) = 6 - 2 = 4 \][/tex]
- Like the previous case, this matches the cost structure for a fractional pound less than 1.
3. Equation 3: [tex]\( f(x) = 6 + 2 \lceil x + 1 \rceil \)[/tex], where [tex]\( x > 0 \)[/tex]
- This equation is not consistent with the definition of the cost structure. Adding 1 before taking the ceil would not correctly reflect the additional cost for each pound or portion thereof.
4. Equation 4: [tex]\( f(x) = 6 + 2 \lfloor x + 1 \rfloor \)[/tex], where [tex]\( x > 0 \)[/tex]
- This equation is also inconsistent. Adding 1 before taking the floor does not correspond to the correctly defined additional costs structure.
### Final Verification:
Let's return to the cost structure implication for multiple precise cases:
- For [tex]\( x = 1 \)[/tex] pound, all equations should yield the same correct cost:
[tex]\[ f(1) = 6 + 2 \times 0 = 6 \][/tex]
- For [tex]\( x = 1.5 \)[/tex] pounds, analyzing the impact further validates that it would fit the ceiling or floor logic correctly, but ceilings are typically more intuitive here since portions of a pound round up.
### Correct Equation:
The correct equation that matches the cost structure:
[tex]\[ f(x) = 6 + 2 \lceil x - 1 \rceil, \; \text{where} \; x > 0 \][/tex]
Thus, the correct function representing the cost is:
[tex]\[ f(x) = 6 + 2\lceil x - 1 \rceil \][/tex]
1. Base Cost and Additional Cost:
- The base cost is [tex]\( \$6 \)[/tex] for packages weighing up to 1 pound.
- For each additional pound or portion of a pound, the cost increases by [tex]\( \$2 \)[/tex].
2. Analyzing the Equations:
We'll evaluate the costs for different values of [tex]\( x \)[/tex], including fractions around 1 pound, to determine which equation fits the given cost structure.
### Equation Analysis
1. Equation 1: [tex]\( f(x) = 6 + 2 \lceil x - 1 \rceil \)[/tex], where [tex]\( x > 0 \)[/tex]
- For [tex]\( x = 0.5 \)[/tex] pounds:
[tex]\[ f(0.5) = 6 + 2 \lceil 0.5 - 1 \rceil = 6 + 2 \lceil -0.5 \rceil = 6 + 2(-1) = 6 - 2 = 4 \][/tex]
- This satisfies the given cost structure since 0.5 would be rounded to 1, resulting in an additional cost.
2. Equation 2: [tex]\( f(x) = 6 + 2 \lfloor x - 1 \rfloor \)[/tex], where [tex]\( x > 0 \)[/tex]
- For [tex]\( x = 0.5 \)[/tex] pounds:
[tex]\[ f(0.5) = 6 + 2 \lfloor 0.5 - 1 \rfloor = 6 + 2 \lfloor -0.5 \rfloor = 6 + 2(-1) = 6 - 2 = 4 \][/tex]
- Like the previous case, this matches the cost structure for a fractional pound less than 1.
3. Equation 3: [tex]\( f(x) = 6 + 2 \lceil x + 1 \rceil \)[/tex], where [tex]\( x > 0 \)[/tex]
- This equation is not consistent with the definition of the cost structure. Adding 1 before taking the ceil would not correctly reflect the additional cost for each pound or portion thereof.
4. Equation 4: [tex]\( f(x) = 6 + 2 \lfloor x + 1 \rfloor \)[/tex], where [tex]\( x > 0 \)[/tex]
- This equation is also inconsistent. Adding 1 before taking the floor does not correspond to the correctly defined additional costs structure.
### Final Verification:
Let's return to the cost structure implication for multiple precise cases:
- For [tex]\( x = 1 \)[/tex] pound, all equations should yield the same correct cost:
[tex]\[ f(1) = 6 + 2 \times 0 = 6 \][/tex]
- For [tex]\( x = 1.5 \)[/tex] pounds, analyzing the impact further validates that it would fit the ceiling or floor logic correctly, but ceilings are typically more intuitive here since portions of a pound round up.
### Correct Equation:
The correct equation that matches the cost structure:
[tex]\[ f(x) = 6 + 2 \lceil x - 1 \rceil, \; \text{where} \; x > 0 \][/tex]
Thus, the correct function representing the cost is:
[tex]\[ f(x) = 6 + 2\lceil x - 1 \rceil \][/tex]
