N1) 1 a) f (x) = 4x³ + 6x+3 ; xo =1.
f '(xo) = f '(1) ---> ?
f '(x) =(4x³ + 6x+3) ' =(4x³ )' + ( 6x) ' +(3) ' =4(x³ )' + 6(x) ' +0 = 4*3*x² +6*1 =12x² +6.
f '(1) =12*1² +6 =12+6 =18.
ответ : 18 .
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д) f(x) = 2x +cos2x ; Xo =π/12 ;
f '(π/12) ---> ?
f ' (x) = (2x +cos2x ) ' =(2x )'+ (cos2x) ' =2(x )'- sin2x*(2x) ' =
2*1 -sin2x*2(x)' =2 - 2sin2x.
f ' (x) =2 - 2sin2x ;
f ' (π/12) =2 - 2sin2*(π/12) =2 - 2sin (π/6) = 2 -2*1/2 =2 -1 =1.
ответ :1 .
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N2) 1. f(x) =x³ +5/2*x² -2x ;
f (X max) ---> ?
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f '(x) = (x³ +5/2*x² -2x )' = (x³) ' +(5/2*x²)'-(2x) ' = 3*x² +5/2*(x²)'-2*(x) ' = 3*x² +5/2*2x -2*1 ;
f '(x) = 3x² +5x -2 ; ***= (x+1)(3x+2) 3(x+1)(x+2/3) *** ;
f '(x) = 0 ;
3x² +5x -2 = 0 ;
D =5² - 4*3*2 =25 - 24 =1 = 1² ;
x₁ = (- 5 - 1)/(2*3) = -1 ;
x₁ = (- 5+ 1)/(2*3) = -2/3 ;
f '(x) = 3(x+1)(x+2/3) ;
f '(x) + - +
---------------------- ( - 1 ) ----------- ( -2/3 )-----------------
f(x) ↑ max ↓ min ↑
x = - 1 _точка максимума ;
f ( -1) = (-1)³ +5/2*( -1)² - 2*(-1) = -1 +2,5 +2 = 3,5.
ответ : 3,5 .
2). f(x) = 1/4*(x^4) - 1/2*x² +5
f(Xmax) --> ?
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f '(x) = ((1/4)*(x^4) - (1/2)*x² +5) ' = ((1/4)*(x^4)) ' - ((1/2)*x²) ' +(5) '
= (1/4)*(x^4) ' - ((1/2)*(x²)) ' + 0 = (1/4)*4X³ -(1/2)*2x =x³ -x ;
f '(x) =x³ -x =x(x² - 1) =x(x-1)(x+1);
f '(x) =(x+1)* x*(x-1) ;
f ' (x) - + - +
--------------- -1 ------------- 0 ---------------- 1 ----------------
f(x) ↓ ↑ max ↓ ↑
x = 0 _точка максимума (Xmax);
f(0) = 1/4*(0)^(4) - (1/2)*(0)² + 5 =5.
ответ : 5 .