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stiki
@stiki
August 2022
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y=x^5+20x^3-65x найдите наибольшее значение функции на отрезке [-3;0]
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sangers1959
Verified answer
Y`=(x⁵+20x³-65x)`=5*x⁴+20*3*x²-65=0 [-3;0]
5x⁴+60x²-65=0
x²=t>0
5t²+60t-65=0 D=4900
t₁=1 t₂=-13 t₂∉
x²=1
x₁=1 x₂=-1
y(-3)=(-3)⁵+20*(-3)³-65*(-3)=-243-540+195=-588.
y(-1)=(-1)⁵+20*(-1)³-65*(-1)=-1-20+65=44=ymax.
y(0)=0⁵+20*0³-65*0=0.
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Answers & Comments
Verified answer
Y`=(x⁵+20x³-65x)`=5*x⁴+20*3*x²-65=0 [-3;0]5x⁴+60x²-65=0
x²=t>0
5t²+60t-65=0 D=4900
t₁=1 t₂=-13 t₂∉
x²=1
x₁=1 x₂=-1
y(-3)=(-3)⁵+20*(-3)³-65*(-3)=-243-540+195=-588.
y(-1)=(-1)⁵+20*(-1)³-65*(-1)=-1-20+65=44=ymax.
y(0)=0⁵+20*0³-65*0=0.