Consider the functions defined implicitly by | vedclass

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(A, D, B) $1.$ Differentiating $y^3-3y+x=0$ with respect to $x$,we get $3y^2y^{\prime}-3y^{\prime}+1=0$,which implies $y^{\prime} = \frac{-1}{3(y^2-1)

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$B, A, D$

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(A, D, B) $1.$ Differentiating $y^3-3y+x=0$ with respect to $x$,we get $3y^2y^{\prime}-3y^{\prime}+1=0$,which implies $y^{\prime} = \frac{-1}{3(y^2-1)}$.At $x = -10\sqrt{2}$,$y = 2\sqrt{2}$,so $y^{\prime} = \frac{-1}{3((2\sqrt{2})^2-1)} = \frac{-1}{3(8-1)} = -\frac{1}{21}$.Differentiating $3y^2y^{\prime}-3y^{\prime}+1=0$ again,we get $6yy^{\prime 2} + 3y^2y^{\prime\prime} - 3y^{\prime\prime} = 0$.$y^{\prime\prime}(3y^2-3) = -6yy^{\prime 2} \Rightarrow y^{\prime\prime} = \frac{-2yy^{\prime 2}}{y^2-1}$.Substituting $y=2\sqrt{2}$ and $y^{\prime}=-\frac{1}{21}$,we get $f^{\prime\prime}(-10\sqrt{2}) = \frac{-2(2\sqrt{2})(-1/21)^2}{8-1} = \frac{-4\sqrt{2}}{7 	imes 441} = -\frac{4\sqrt{2}}{7^3 	imes 3^2}$.$2.$ The area is $\int_a^b |f(x)| dx$. Since $f(x) Using integration by parts: $\int f(x) dx = xf(x) - \int xf^{\prime}(x) dx$.Since $f^{\prime}(x) = \frac{-1}{3(f(x)^2-1)}$,the integral is $bf(b)-af(a) - \int_a^b x \left(\frac{-1}{3(f(x)^2-1)}ight) dx = bf(b)-af(a) + \int_a^b \frac{x}{3(f(x)^2-1)} dx$.The area is $-\int_a^b f(x) dx = -bf(b)+af(a) - \int_a^b \frac{x}{3(f(x)^2-1)} dx$.$3.$ $\int_{-1}^1 g^{\prime}(x) dx = g(1) - g(-1)$.Since $g(x)^3 - 3g(x) + x = 0$,$g(-x)^3 - 3g(-x) - x = 0$. Let $h(x) = -g(-x)$,then $(-h(x))^3 - 3(-h(x)) - x = 0 \Rightarrow -h(x)^3 + 3h(x) - x = 0 \Rightarrow h(x)^3 - 3h(x) + x = 0$.Thus $g(x) = -g(-x)$,so $g$ is an odd function. $g(-1) = -g(1)$.Therefore,$g(1) - g(-1) = g(1) - (-g(1)) = 2g(1)$.

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