Question

Найдите дельта x и дельта f в точке с абсциссой x0 и отношение дельта x/дельта f
1)f(x) = sin3x - 2 x0=п/3; х=п/4
2)f(x) = cos2x + 2 x0=-п/3; х=-п/4

Answer 1

$f(x) = \sin3x - 2$

$\Delta x=x-x_0\\\Delta x=\dfrac{\pi}{4} -\dfrac{\pi}{3} =\dfrac{3\pi}{12} -\dfrac{4\pi}{12}=-\dfrac{\pi}{12}$

$\Delta f=f(x)-f(x_0)\\\Delta f=\left(\sin\left(3\cdot\dfrac{\pi}{4} \right)-2\right)-\left(\sin\left(3\cdot\dfrac{\pi}{3} \right)-2\right)=\\=\sin\dfrac{3\pi}{4} -2-\sin\pi+2=\dfrac{\sqrt{2} }{2}-0=\dfrac{\sqrt{2} }{2}$

$\dfrac{\Delta x}{\Delta f} =-\dfrac{\pi }{12} :\dfrac{\sqrt{2} }{2} =-\dfrac{\pi }{12} \cdot\dfrac{2 }{\sqrt{2}}=-\dfrac{\pi }{12} \cdot\sqrt{2}=-\dfrac{\pi\sqrt{2}}{12}$

$f(x) = \cos2x + 2$

$\Delta x=x-x_0\\\Delta x=-\dfrac{\pi}{4} -\left(-\dfrac{\pi}{3}\right) =-\dfrac{3\pi}{12} +\dfrac{4\pi}{12}=\dfrac{\pi}{12}$

$\Delta f=f(x)-f(x_0)\\\Delta f=\left(\cos\left(2\cdot\left(-\dfrac{\pi}{4}\right) \right)+2\right)-\left(\cos\left(2\cdot\left(-\dfrac{\pi}{3}\right) \right)+2\right)=\\=\cos\dfrac{\pi}{2} +2-\cos\dfrac{2\pi}{3}-2=0-\left(-\dfrac{1}{2}\right)-0=\dfrac{1}{2}$

$\dfrac{\Delta x}{\Delta f} =\dfrac{\pi }{12} :\dfrac{1}{2} =\dfrac{\pi }{12}\cdot2=\dfrac{\pi }{6}$