Задача 30015 Найти наименьшее значение функции y =...
Условие
y = √x2–2x+2+√x2–10x+29 [л12]
Решение
x2–2x+2=(x–1)2+1 > 0 при любом х ∈ (– ∞ ; + ∞ )
x2–10x+29=(x–5)2+4 > 0 при любом х ∈ (– ∞ ; + ∞ )
Область определения функции (– ∞ ; + ∞ )
y`= (√x2–2x+2 + √x2–10x+29)` =
=(2x–2)/(2·√x2–2x+2) + (2x–10)/(2·√x2–10x+29) =
=(x–1)/√x2–2x+2 + (x–5)/√x2–10x+29
y`=0
(x–1)/√x2–2x+2 + (x–5)/√x2–10x+29 = 0
(x–1)·√(x–5)2+4+(x–5)·√(x–1)2+1=0
(x–1)·√(x–5)2+4= –(x–5)·√(x–1)2+1
(x–1)·√(x–5)2+4= (5–х)·√(x–1)2+1
{x–1 ≥ 0;
{5–x ≥ 0;
{(x–1)2·((x–5)2+4)=(5–x)2·((x–1)2+1).
{1 ≤ x ≤ 5;
{4·(x–1)2=(5–x)2.
4x2–8x+4=25–10x+x2
3x2+2x–21=0
D=4–4·3·(–21)=4+252=256
x1=(–2–16)/6=–3 ∉ [1;5] или x2=(–2+16)/6= 7/3
х=7/3 – точка минимума, производная меняет знак с – на +
[1] __–__ (7/3) ___+____ [5]
ннаименьшее=y(7/3)=√((7/3)–1)2+1 + √((7/3)–5)2+4)=
=√(4/3)2+1 + √(–8/3)2+4=
=√25/9 +√100/9=(5/3)+(10/3)=15/3 = 5
О т в е т. 5