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Home Question Bank Math Calculus Archive 2024 November 26

Calculus Questions from Nov 26,2024

Browse the Calculus Q&A Archive for Nov 26,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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What are the horizontal and vertical asymptotes of the function \( f(x)=\frac{1}{x}+7 \) ? Horizontal: Vertical: Paso 2. Calcular la pendiente \( F^{\prime}(x)=-\mathrm{m} \) What are the horizontal and vertical asymptotes of the function \( f(x)=\frac{1}{x}-5 \) ? Horizontal: Vertical: Determinar la longitud de arco de la hélice circular con función vectorial \( r(t)=\cos (t) i+\operatorname{sen}(t) j+t k \) del punto \( (1,0,0) \) al punto \( (1,0,2 \pi) \) Calculé la siguiente integral por partes \( \int x \ln (x) d x \) (A) \( \frac{x^{2}}{2} \ln (x)-\frac{1}{4} x^{2}+C \) (B) \( 2 \frac{x^{2}}{3} \ln (x)-\frac{3}{4} x^{3}+C \) (C) \( \frac{x^{2}}{2} \ln (x)+\frac{1}{8} x^{2}+C \) (D) \( \frac{x}{2} \ln (x)-\frac{1}{3} x^{3}+C \) Calcule la siguiente integral por partes \( \int x \ln (x) d x \) (A) \( \frac{x^{2}}{2} \ln (x)-\frac{1}{4} x^{2}+C \) (B) \( 2 \frac{x^{2}}{3} \ln (x)-\frac{3}{4} x^{3}+C \) (C) \( \frac{x^{2}}{2} \ln (x)+\frac{1}{8} x^{2}+C \) (D) \( \frac{x}{2} \ln (x)-\frac{1}{3} x^{3}+C \) 1) Calcular los máximos y mínimos, así como los puntos de inflexión de la función definida como \( f(x)=x^{3}-3 x+2 \) y trazar su gráfica. \( \begin{array}{l}x=16 \cos t, y=1 \sin t, t=-\frac{-}{4} \\ \text { Find } \frac{d y}{d x} \\ \frac{d y}{d x}=\frac{4 \cos t}{-16 \sin t} \\ \text { Find the slope of the tangent line at } t=-\frac{\pi}{4} \text {. } \\ \text { The slope is } \frac{1}{4} \text {. } \\ \text { The equation } \square \text { represents the line tangent to the curve at } t=-\frac{\pi}{4} \text {. } \\ \text { (Type an exact answer, using radicals as needed.) }\end{array} \). Find an equation for the line tangent to the curve at the point defined by the given value of \( t \) Also, find the value of \( \frac{d^{2} y}{d x^{2}} \) at this point. \( x=16 \cos t, y=4 \sin t, t=-\frac{\pi}{4} \) \( \frac{d y}{d x}=\frac{d y}{d x} \). \( 16 \sin t \) Find the slope of the tangent line at \( t=-\frac{\pi}{4} \) The slope is \( \operatorname { Lim } \frac { \sqrt { x ^ { 2 } + 9 } } { x } \)
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