Ответы
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task/29744057 ---------------------
V₁(t) =S₁'(t) =(2t³ -4t²+5t) '=2*3t²-4*2t +5 =6t² -8t+5.⇒ a₁(t)=V₁'(t)= (6t²-8t)' =12t-8.
V₂(t) =S₂'(t) =(2t³ -1,5t²) ' =2*3t² -1,5*2t = 6t² -3t . ⇒ a₂(t)=V₂'(t)= (6t² -3t)' =12t -3.
V₁(t₀) = V₂(t₀ ) ⇔ 6t₀² - 8t₀ +5 = 6t₀² -3t₀ ⇒ t₀=1 (c)
a₁(t₀)=12t₀ -8 =12*1 - 8 =0 = 4 ( м/с²) ; a₂(t₀) = 12*1 -3 =9 ( м/с² ) .
* * * * * * *
f(x) =(x²+8) / (x+1) ; x ∈[ 0 ;3] . max f(x) - ? ; min f(x) -?
f '(x) = ( (x²+8) / (x+1) ) ' = [ (x²+8)'*(x+1) - (x²+8) *(x+1) ' ] /(x+1)² =
= [ 2x(x+1) - (x²+8) ] *(x+1)² =(x² +2x -8) /(x+1)² = (x+4)(x-2) / (x+1)²
* f(x) = ( x²+8) / (x+1) =(x² - 1) / (x+1) +9 /(x+1) = x -1 + 9*( x+1 )⁻¹ ; f '(x) =1 - 9 / (x+1)² = [1 -3/(x+1) ]*[1+ 3/(x+1) ] =(x-2)(x+4) / (x+1)² * * *
критические (стационарные) точки :
f ' (x) = 0 ; (x-2)(x+4) / (x+1)² =0 ⇒ [ x = 2 ; x = - 4 ∉ [ 0 ;3 ] .
f(0) =(0²+8) /(0+1) = 8 ; f(3) =(3²+8)/(3+1) =17/4 =4,25 ; f(2) =(2²+8)/(2+1) = 4 .
max f(x) = max {8 ; 4,25; 4} = 8 .
min f(x) = min {8 ; 4,25; 4} = 4 .
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