Question

Производная. Решите все три задания.​

Answer 1

$1.\,k=\frac{\sqrt2}{2};\\\\2.\,y_k=x-1\\\\3.\; x_{01}=0,\,x_{02}=6$

Пошаговое объяснение:

$1.\ y=-\frac{1}{3}sin3x+2cos4x-3,\,x_0=\frac{\pi}{4},\, k-?\\\\y'=(-\frac{1}{3}sin3x+2cos4x-3)'=-\frac{1}{3}(sin3x)'+2(cos4x)'-(3)'=\\\\=-\frac{1}{3}cos3x\cdot(3x)'+2(-sin4x)\cdot(4x)'-0=-\frac{1}{3}cos3x\cdot3-2sin4x\cdot4=\\\\=-cos3x-8sin4x\\\\k=y'(x_0)=-cos(3\cdot\frac{\pi}{4})-8sin(4\cdot\frac{\pi}{4})=-cos(\frac{3\pi}{4})-8sin\pi=-(-\frac{\sqrt2}{2})-8\cdot0=\frac{\sqrt2}{2}$

$2.\,f(x)=x^3+3x^2+4x;\;x_0=-1;\;y_k-?\\\\y_k=f(x_0)+f'(x_0)(x-x_0)\\\\f'(x)=(x^3+3x^2+4x)'=(x^3)'+3(x^2)'+4(x)'=3x^2+6x+4\\\\y_k=(x_0^3+3x_0^2+4x_0)+(3x_0^2+6x_0+4)(x-x_0)=\\\\=((-1)^3+3\cdot(-1)^2+4\cdot(-1))+(3\cdot(-1)^2+6\cdot(-1)+4)(x-(-1))=\\\\=(-1+3-4)+(3-6+4)(x+1)=-2+x+1=x-1$

$3.\;f(x)=\frac{1}{3}x^3-3x^2-x-6;\;\alpha=135^{\circ};\; x_0-?\\\\f'(x_0)=k=tg(\alpha) \Rightarrow f'(x_0)=x_0^2-6x_0-1=tg(135^{\circ})\\\\x_0^2-6x_0-1=-1\\\\x_0^2-6x_0=0\\\\x_0(x_0-6)=0\Rightarrow x_{01}=0;\;x_{02}=6$