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

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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). $$ P(x)=\frac{11 x}{1-x} \quad 0 \leq x<1 $$ (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 concave upward on $$ \square $$ and concave downward on B. $$ P(x) $$ is concave upward on $$ [0,1) $$. C. $$ P(x) $$ is concave downward on (C) Find any horizontal and vertical asymptote. Find the vertical asymptote(s), if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The vertical asymptote(s) is (are) $$ x= $$ $$ \square $$ (Simplify your answer. Use a comma to separate answers as needed) B. There are no vertical asymptotes.

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To solve the problem, we will analyze the function $$ P(x) = \frac{11x}{1-x} $$ step by step. ### Part A: Determine Concavity 1. **Find the first derivative $$ P'(x) $$**: $$ P'(x) = \frac{(1-x)(11) - 11x(-1)}{(1-x)^2} = \frac{11(1-x) + 11x}{(1-x)^2} = \frac{11}{(1-x)^2} $$ 2. **Find the second derivative $$ P''(x) $$**: $$ P''(x) = \frac{d}{dx}\left(\frac{11}{(1-x)^2}\right) = 11 \cdot \frac{d}{dx}((1-x)^{-2}) = 11 \cdot (-2)(1-x)^{-3}(-1) = \frac{22}{(1-x)^3} $$ 3. **Determine the sign of $$ P''(x) $$**: - Since $$ (1-x)^3 > 0 $$ for $$ 0 \leq x < 1 $$, $$ P''(x) > 0 $$ in this interval. - Therefore, $$ P(x) $$ is concave upward on $$ [0, 1) $$. ### Part B: Concave Downward Intervals Since $$ P''(x) > 0 $$ for $$ 0 \leq x < 1 $$, there are no intervals where $$ P(x) $$ is concave downward. Thus, we can conclude: - **Concave downward on**: There are no intervals. ### Part C: Asymptotes 1. **Vertical Asymptote**: - A vertical asymptote occurs where the denominator is zero. Set $$ 1 - x = 0 $$: $$ x = 1 $$ - Therefore, the vertical asymptote is at $$ x = 1 $$. 2. **Horizontal Asymptote**: - To find the horizontal asymptote, we analyze the limit as $$ x $$ approaches 1: $$ \lim_{x \to 1} P(x) = \lim_{x \to 1} \frac{11x}{1-x} = \infty $$ - There is no horizontal asymptote since the function approaches infinity. ### Final Answers A. $$ P(x) $$ is concave upward on $$ [0, 1) $$ and concave downward on $$ \emptyset $$. B. $$ P(x) $$ is concave downward on $$ \emptyset $$. C. The vertical asymptote(s) is (are) $$ x = 1 $$. Thus, the answers are: - A: $$ [0, 1) $$ - B: $$ \emptyset $$ - C: $$ x = 1 $$ A. $$ P(x) $$ is concave upward on $$ [0, 1) $$ and concave downward on none. B. $$ P(x) $$ is concave downward on none. C. The vertical asymptote is at $$ x = 1 $$.

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