Matematika - LS 24/25

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Určete Taylorův polynom druhého stupně funkce

f(x)=\sin(3x+\frac{\pi}{6})+2. se středem v bodě a=\frac{\pi}{6}.

2+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2

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2+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

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2+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{6})-\frac{\sqrt 3}{4}(x-\frac{\pi}{6})^2

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2+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

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2+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x+\frac{\pi}{6})^2

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\frac{3}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{9}{4}(x-\frac{\pi}{6})^2

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Nalezněte obecné řešení rovnice

y'y^{8}-\frac{1}{e^{{-x}}}=0.

y=\sqrt[9]{8+e^x+C}

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y=\sqrt[9]{9(e^x+C)}

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Žádná z předchozích odpovědí není správně.

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y=\sqrt[7]{7(-e^x+C)}

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y=\sqrt[9]{9(e^x)}+C

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Určete obsah rovinného obrazce ohraničeného křivkami x-y=2 a y=x^{2}-3x-2.

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Spočtěte \int_{-7\pi}^{-5\pi}\limits \cos \frac{x-4\pi}{6}\ \mathrm{d}x

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Určete Taylorův polynom druhého stupně funkce

f(x)=\cos(5x-\pi)+6. se středem v bodě a=\frac{\pi}{3}.

\frac{11}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{2}(x-\frac{\pi}{3})^2

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\frac{11}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{4}(x-\frac{\pi}{3})^2

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\frac{11}{2}-\frac{\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{1}{4}(x-\frac{\pi}{3})^2

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\frac{11}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{4}(x-\frac{\pi}{3})^2

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\frac{11}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{2}(x-\frac{\pi}{3})^2

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\frac{11}{2}-\frac{5\sqrt 3}{2}(x+\frac{\pi}{3})-\frac{25}{2}(x+\frac{\pi}{3})^2

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Spočtěte neurčitý integrál

\displaystyle\int\frac{x\cdot 3^x+1}{4x}dx.

\displaystyle\frac{3^x+2(x-1)}{4(x-1)}+C

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\displaystyle\frac{3^x+x\cdot 3^x\cdot \ln3}{4}+C

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\displaystyle\frac{3^x+\ln3\cdot \ln |x|}{4\ln 3}+C

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\displaystyle\frac{x\cdot 3^x+2\ln3}{4x\cdot \ln 3}+C

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\displaystyle\frac{(x-1)3^x+x}{2x^2}+C

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\displaystyle\frac{x^2\cdot 3^x\cdot \ln3-1}{4x^2}+C

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