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Differentiation is a mathematical process used to find the rate at which a certain quantity changes. When dealing with functions, like our cost function for purifying water, differentiation helps us discover how small changes in one quantity (purity percentage in this case) affect another quantity (cost).

More technically, to differentiate a function, you need to find its derivative. For our cost function \( C(x) = \frac{100}{100-x} \), we wanted to know the instantaneous rate of change of cost with respect to the purity percentage. To achieve this, we used the differentiation rule for a reciprocal function \( f(x) = \frac{1}{g(x)} \), which delivers the derivative:

• \( f'(x) = -\frac{g'(x)}{g(x)^2} \)

For our specific function \( C(x) \), we set \( g(x) = 100-x \) and found \( g'(x) = -1 \). Substituting these into the formula gives the derivative:

• \[ C'(x) = \frac{100}{(100-x)^2} \]

This derivative \( C'(x) \) tells us how quickly the cost changes as the purity increases.

Cost functions describe how much financial cost is associated with a specific activity or level of production. In the context of water purification, the cost function \( C(x) = \frac{100}{100-x} \) calculates the expense in cents needed to achieve a particular purity percentage \( x \) for a gallon of water.

The domain of the function is limited to \( 50 \leq x < 100 \), implying that purification below 50% isn’t considered, and 100% purity is theoretically infinite, hence not practical.

By evaluating this function at various points, we gain insights into how costs escalate. For example:

• At 95% purity, the derivative \( C'(95) = 4 \) reveals that the cost increases by 4 cents for each additional percent of purity achieved.
• At 98% purity, \( C'(98) = 25 \) shows a dramatic increase, indicating that nearing complete purity becomes disproportionately expensive.

The analysis of this function and its derivative can assist businesses in decision-making concerning the cost-effectiveness of reaching higher purity standards.

Purity percentage is a measure of how clean or pure a substance is, typically presented as a percentage. In our exercise, it's the percentage of purity achieved in the water purification process.

Understanding how purity levels impact costs is essential, especially in industries focused on high purity standards. As the function \( C(x) = \frac{100}{100-x} \) shows, achieving higher purity levels becomes increasingly costly as \( x \) approaches 100%.

By calculating the derivative, we learn:

• At 95% purity, the cost rise is moderate (4 cents per percent), reflecting manageable expenses for high, but not extreme, purity levels.
• As purity reaches 98%, costs escalate dramatically (25 cents per percent), highlighting the challenges and expense of reaching near-perfect purity.

This steep increase near full purity underscores the exponential costs and potential fruitlessness of totally pure products, guiding businesses in determining practical and financial purity targets.