Advanced Placement (AP), 20.09.2020 08:01 HourlongNine342

Four students are at an extracurricular activity fair at their high school and are trying to decide which clubs to join. Some information about the students is listed in this chart: A 2-column table with 4 rows. Column 1 is labeled Student with entries Eleanor, Russel, Giada, Joe. Column 2 is labeled Hobbies with entires is already in B P A but is looking for a similae club to join as well; enjoys being treasurer for school band and drama club, and wants to join a club to help him learn more about these skills; builds remote-control cars and robots as a hobby; wants to be a stay-at-home dad and be involved in the homeowners association for his future. Which describes which CTSO each student should join?

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Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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Find the area of the region bounded by the curves y equals the inverse sine of x divided by 4, y = 0, andx = 4 obtained by integrating with respect to y. your work must include the definite integral and theantiderivative.

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