Помогите пожалуйста решить. Created by kristi1998. algebra-ru

Помогите пожалуйста решить...

natylia64 1.
а) f(x) =(1/2)*x⁴ -x⁵ +5 ;
f '(x) = ((1/2)*x⁴ -x⁵ +5)= (1/2)*4*x³ -5x⁴ +0 =2x³ -5x⁴.
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б) f(x)=4x -1/x³.
f'(x)=(4x -1/x³) ' =(4x - x^(-3)) '= 4 + 3/x⁴.
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2.
а) f(x)=x*cosx
f'(x)= (x*cosx)' =(x)'*cosx+x*(cosx)' =cosx-x*sinx.
f'(-π/2)= cos(-π/2)- (-π/2)*sin(-π/2) =cosπ/2 +π/2*(-sinπ/2)= - π/2.
б) f(x) =(3x+4)⁵ ;
f'(x) = ((3x+4)⁵)' =5(3x+4)⁴*(3x+4)'=5(3x+4)⁴*3=15(3x+4)⁴.
f'(-1) =15(3*(-1)+4)⁴ =15.
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3.
f(x) =cos2x +√3*x ; f '(x)=0 . ||ошибочно f '(-1)=0 ???? ||
f '(x) =(cos2x +√3*x)' = -2sin2x +√3 .
f '(x) =0 ⇔ -2sin2x +√3 =0⇒sin2x =(√3)/2 ;
2x =(-1)^n*π/3 +πn , n∈Z.
x =(-1)^n*π/6 +π*n/2 , n∈Z.
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4.
f(x) =6x -x³ ; f '(x) ≤0.
f '(x) =(6x -x³)' =6 -3x² =3(2 -x²).
f'(x) ≤ 0 ⇒ 3(2 -x²) ≤ 0 ⇔ x² -2 ≥0 ⇒x∈(-∞; -√2] ∪[√2;∞).

1 votes Thanks 2

oganesbagoyan

Verified answer

1)а) f(x)= 1/2x^4-x^5+5= (1/2x^4)'-(x^5)'+(5)'=1/2*4x^3-5x^4=2x^3-5x^4
б) f(x)=4x-1/x^3=(4x)'-(1/x^3)'=4-(x^-3)'=4+3/x^4

2 votes Thanks 1

kristi1998 Спасибо большое

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