In Silicon Valley, a number of computer firms were found to be contaminating underground water supplies with toxic chemicals stored in leaking underground containers. A water quality control agency ordered the companies to take immediate corrective action and contribute to a monetary pool for the testing and cleanup of the underground contamination. Suppose that the required monetary pool (in millions of dollars) is given by the following function, where \( x \) is the percentage (expressed as a decimal fraction) of the total contaminant removed. Complete parts (A) through (E). \( P(x)=\frac{11 x}{1-x} \quad 0 \leq x<1 \) (A) Where is \( P(x) \) increasing? Decreasing? Select the correct choice below and fill in the answer box(es) to complete your choice. (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. \( P(x) \) is increasing on \( \square \) and decreasing on B. \( P(x) \) is increasing on C. \( P(x) \) is decreasing on \( \square \).
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To determine where \( P(x) \) is increasing or decreasing, we first need to find the derivative of \( P(x) \). Calculating the derivative of \( P(x) = \frac{11x}{1-x} \) using the quotient rule yields: \[ P'(x) = \frac{(1-x)(11) - 11x(-1)}{(1-x)^2} = \frac{11(1-x) + 11x}{(1-x)^2} = \frac{11}{(1-x)^2} \] This derivative \( P'(x) \) is positive for all \( x \) in the interval \( [0, 1) \) since \( 11 \) is positive and \( (1-x)^2 \) is always positive (as long as \( x < 1 \)). Therefore, \( P(x) \) is always increasing on this interval. Now filling in the answers: A. \( P(x) \) is increasing on \( [0, 1) \) and decreasing on ∅. B. \( P(x) \) is increasing on \( [0, 1) \). C. \( P(x) \) is decreasing on ∅.
