Đặt A=\(\frac{3-4x}{x^2+1}\)
*Tìm GTNN:
A = \(\frac{3-4x}{x^2+1}\) = \(\frac{x^2-4x+4-x^2-1}{x^2+1}=\frac{\left(x-2\right)^2-\left(x^2+1\right)}{x^2+1}\) = \(\frac{\left(x-2\right)^2}{x^2+1}-1\)
Vì \(\frac{\left(x-2\right)^2}{x^2+1}\ge0\) ∀ x => \(\frac{\left(x-2\right)^2}{x^2+1}-1\) ≥ -1 ∀ x hay A ≥ -1 ∀ x
Dấu "=" xảy ra ⇔ x - 2 = 0 ⇔ x = 2
Vậy minA = -1 ⇔ x = 2
*Tìm GTLN:
A = \(\frac{3-4x}{x^2+1}\) = \(\frac{-4x^2-4x-1+4x^2+4}{x^2+1}=\frac{-\left(2x+1\right)^2+4\left(x^2+1\right)}{x^2+1}\)=\(\frac{-\left(2x+1\right)^2}{x^2+1}+4\)
Vì \(\frac{-\left(2x-1\right)^2}{x^2-1}\) ≤ 0 ∀ x => \(\frac{-\left(2x+1\right)^2}{x^2+1}+4\) ≤ 4 ∀ x hay A ≤ 4 ∀ x
Dấu "=" xảy ra ⇔ 2x + 1 = 0 ⇔ 2x = -1 ⇔ x = \(\frac{-1}{2}\)
Vậy maxA = 4 ⇔ x = \(\frac{-1}{2}\)
