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Matematika - LS 24/25

Matematika - LS 24/25

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Určete Taylorův polynom druhého stupně funkce

$f(x)=\sin(5x-\frac{\pi}{6})-1.$ f(x)=\sin(5x-\frac{\pi}{6})-1. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

$-1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})+\frac{25\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})+\frac{25\sqrt 3}{4}(x-\frac{\pi}{6})^2

$-1+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})-\frac{25\sqrt 3}{2}(x+\frac{\pi}{6})^2$ -1+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{6})-\frac{25\sqrt 3}{2}(x+\frac{\pi}{6})^2

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

$-1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})-\frac{25\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})-\frac{25\sqrt 3}{4}(x-\frac{\pi}{6})^2

$-1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})-\frac{25\sqrt 3}{2}(x-\frac{\pi}{6})^2$ -1+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{6})-\frac{25\sqrt 3}{2}(x-\frac{\pi}{6})^2

$\frac{-3}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{25}{4}(x-\frac{\pi}{6})^2$ \frac{-3}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{25}{4}(x-\frac{\pi}{6})^2

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Nalezněte obecné řešení rovnice

$y'y^{5}-\frac{1}{e^{{-x}}}=0.$ y'y^{5}-\frac{1}{e^{{-x}}}=0.

$y=\sqrt[6]{\frac{(e^x+C)}{6}}$ y=\sqrt[6]{\frac{(e^x+C)}{6}}

$y=\sqrt[6]{6(e^x+C)}$ y=\sqrt[6]{6(e^x+C)}

Žádná z předchozích odpovědí není správně.

$y=\sqrt[6]{5+e^x+C}$ y=\sqrt[6]{5+e^x+C}

$y=\sqrt[6]{6(e^x)}+C$ y=\sqrt[6]{6(e^x)}+C

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Máme funkci f spojitou a sudou na celém R a víme, že $\int_0^a\limits f(x)\ \mathrm{d}x=P$ \int_0^a\limits f(x)\ \mathrm{d}x=P. Pak $\int_{-a}^a\limits f(x)\ \mathrm{d}x$ \int_{-a}^a\limits f(x)\ \mathrm{d}x je roven

-P

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Spočtěte obecné řešení diferenciální rovnice y'+4x^-1y=x⁴.

Žádná z odpovědí a–d není správně.

y=-4/5 x⁵+ln|x|+C

y=x^-4+Cx⁴

y=1/9 x⁵+Cx^-4

y=x⁴+Cx^-4

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Spočtěte neurčitý integrál

$\displaystyle\int\left(\frac{1}{x^{8}}+e\right)dx.$ \displaystyle\int\left(\frac{1}{x^{8}}+e\right)dx.

$\displaystyle\frac{9}{x^{9}}+ex+C$ \displaystyle\frac{9}{x^{9}}+ex+C

$\displaystyle\frac{9}{x^{9}}+e^x+C$ \displaystyle\frac{9}{x^{9}}+e^x+C

$\displaystyle\frac{-1}{7x^{7}}+e^x+C$ \displaystyle\frac{-1}{7x^{7}}+e^x+C

$\displaystyle\ln\left|x^{8}\right|+ex+C$ \displaystyle\ln\left|x^{8}\right|+ex+C

$\displaystyle\frac{-1}{7x^{7}}+ex+C$ \displaystyle\frac{-1}{7x^{7}}+ex+C

$\displaystyle\ln\left|x^{8}\right|+e^x+C$ \displaystyle\ln\left|x^{8}\right|+e^x+C

❌

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Určete Taylorův polynom druhého stupně funkce

$f(x)=\sin(5x-\pi)-3.$ f(x)=\sin(5x-\pi)-3. se středem v bodě $a=\frac{\pi}{4}$ a=\frac{\pi}{4}.

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

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

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

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

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

$-3+\frac{\sqrt 2}{2}+\frac{5\sqrt 2}{2}(x-\frac{\pi}{4})+\frac{25\sqrt 2}{4}(x-\frac{\pi}{4})^2$ -3+\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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Spočtěte $\displaystyle \int_1^{e^2} \frac{1}{x}\ \mathrm{d}x$ \displaystyle \int_1^{e^2} \frac{1}{x}\ \mathrm{d}x (zaokrouhlete na dvě desetinná místa).