Matematika - LS 24/25

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Spočtěte \int_{4\pi}^{6\pi}\limits \sin \frac{x+6\pi}{4}\ \mathrm{d}x

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Určete Taylorův polynom druhého stupně funkce

f(x)=\cos(5x-\frac{\pi}{2})+9. se středem v bodě a=\frac{\pi}{4}.

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

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

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

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

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

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

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Nalezněte obecné řešení rovnice

e^{{-y}}-y'x^{7}=0

y=e^{\frac{x^{-6}}{-6}+C}

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y=\ln (\frac{x^{8}}{8}+C)

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y=\frac{\ln (x^{-6})}{-6}+C

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y=\ln (\frac{x^{-6}}{-6}+C)

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Žádná z předchozích odpovědí není správně.

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Spočtěte \int_{-7\pi}^{-5\pi}\limits \sin \frac{x+2\pi}{3}\ \mathrm{d}x

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

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Nalezněte obecné řešení rovnice

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

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

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

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

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

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Žádná z předchozích odpovědí není správně.

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