Given that f ′′ ( x ) = cos ( x ) f ″ ( x ) = cos ⁡ ( x ) , f ′ ( π / 2 ) = 7 f ′ ( π / 2 ) = 7 and f ( π / 2 ) = 5 f ( π / 2 ) = 5 find: f

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Given that f ′′ ( x ) = cos ( x ) f ″ ( x ) = cos ⁡ ( x ) , f ′ ( π / 2 ) = 7 f ′ ( π / 2 ) = 7 and f ( π / 2 ) = 5 f ( π / 2 ) = 5 find: f ′ ( x ) = f ′ ( x ) = f ( x ) = f ( x ) =

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Khải Quang 4 years 2021-08-13T04:28:32+00:00 1 Answers 13 views 0

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    2021-08-13T04:29:48+00:00

    Answer: \sin x+6

    -\cos x+6x+5-3\pi

    Step-by-step explanation:

    Given

    f''(x)=\cos x

    Integrating the equation

    \Rightarrow f''(x)=\dfrac{d^2y}{dx^2}=\cos x\\\\\Rightarrow \int \frac{\mathrm{d^2} y}{\mathrm{d} x^2}=\int \cos x\\\\\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\sin x+c

    Put x=\frac{\pi}{2}\ \text{and}\ \frac{dy}{dx}=7

    \Rightarrow 7=\sin \frac{\pi }{2}+c\\\Rightarrow c=7-1\\\Rightarrow c=6

    \Rightarrow \dfrac{dy}{dx}=\sin x+6

    again integrating both sides of the equation

    \Rightarrow \int \frac{\mathrm{d} y}{\mathrm{d} x}=\int (\sin x+6)dx\\\\\Rightarrow y=-\cos x+6x+c_1

    Put  x=\frac{\pi}{2}\ \text{and}\ y=5

    \Rightarrow 5=-\cos \frac{\pi}{2}+6(\frac{\pi}{2})+c_1\\\Rightarrow c_1=5-3\pi

    \therefore f(x)=y=-\cos x+6x+5-3\pi

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