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

$f(x) = sin3x - 2$

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

$Delta f=f(x)-f(x_0)\Delta f=left(sinleft(3cdotdfrac{pi}{4} right)-2right)-left(sinleft(3cdotdfrac{pi}{3} right)-2right)=\=sindfrac{3pi}{4} -2-sinpi+2=dfrac{sqrt{2} }{2}-0=dfrac{sqrt{2} }{2}$

$dfrac{Delta x}{Delta f} =-dfrac{pi }{12} :dfrac{sqrt{2} }{2} =-dfrac{pi }{12} cdotdfrac{2 }{sqrt{2}}=-dfrac{pi }{12} cdotsqrt{2}=-dfrac{pisqrt{2}}{12}$

$f(x) = cos2x + 2$

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

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

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