\(H=2x^2-x+4==2\left(x^2-\frac{1}{2}x+2\right)\)
\(=2\left[x^2-2\cdot x\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2\right]+\frac{31}{8}\)
\(=2\left(x-\frac{1}{4}\right)^2+\frac{31}{8}\)
Vì \(\left(x-\frac{1}{4}\right)^2\ge0\forall x\)
=> \(2\left(x-\frac{1}{4}\right)^2+\frac{31}{8}\ge\frac{31}{8}\forall x\)
Dấu " = " xảy ra khi và chỉ khi \(\left(x-\frac{1}{4}\right)^2=0\Rightarrow x=\frac{1}{4}\)
Vậy \(H_{min}=\frac{31}{8}\)khi x = 1/4
2) \(I=\frac{1}{2}x^2+3x=\frac{1}{2}\left(x^2+6x\right)\)
\(=\frac{1}{2}\left(x^2+2\cdot x\cdot3+3^2\right)-\frac{9}{2}\)
\(=\frac{1}{2}\left(x+3\right)^2-\frac{9}{2}\)
Vì \(\left(x+3\right)^2\ge0\forall x\)
=> \(\frac{1}{2}\left(x+3\right)^2-\frac{9}{2}\ge-\frac{9}{2}\forall x\)
Dấu " = " xảy ra khi và chỉ khi (x + 3)2 = 0 => x = -3
Vậy \(I_{min}=-\frac{9}{2}\)khi x = -3
1) \(H=2x^2-x+4=2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{31}{8}=2\left(x-\frac{1}{4}\right)^2+\frac{31}{8}\ge\frac{31}{8}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(2\left(x-\frac{1}{4}\right)^2\ge0\Rightarrow x=\frac{1}{4}\)
Vậy Min(H) = 31/8 khi x = 1/4
2) \(I=\frac{1}{2}x^2+3x=\frac{1}{2}\left(x^2+6x+9\right)-\frac{9}{2}=\frac{1}{2}\left(x+3\right)^2-\frac{9}{2}\ge-\frac{9}{2}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\frac{1}{2}\left(x+3\right)^2=0\Rightarrow x=-3\)
Vậy Min(I) = -9/2 khi x = -3
H = 2x2 - x + 4 = 2( x2 - 1/2x + 1/16 ) + 31/8 = 2( x - 1/4 )2 + 31/8 ≥ 31/8 ∀ x
Dấu "=" xảy ra khi x = 1/4
=> MinH = 31/8 <=> x = 1/4
I = 1/2x2 + 3x = 1/2( x2 + 6x + 9 ) - 9/2 = 1/2( x + 3 )2 - 9/2 ≥ -9/2 ∀ x
Dấu "=" xảy ra khi x = -3
=> MinI = -9/2 <=> x = -3
