Question

Tìm GTLN và GTNN của \(\frac{4x}{4x^2+1}\)

Answer 1

\(\frac{4x}{4x^2+1}=\frac{x}{x^2+\frac{1}{4}}=\frac{\left(x^2+\frac{1}{4}\right)-\left(x^2-x+\frac{1}{4}\right)}{x^2+\frac{1}{4}}=1-\frac{\left(x-\frac{1}{2}\right)^2}{x^2+\frac{1}{4}}\le1\)dấu "=" xảy ra <=> \(x-\frac{1}{2}=0\) <=>x=\(\frac{1}{2}\)

vậy max=1 khi x=1/2

Answer 2

\(\frac{4x}{4x^2+1}=\frac{x}{x^2+\frac{1}{4}}=\frac{\left(x^2+x+\frac{1}{4}\right)-\left(x^2+\frac{1}{4}\right)}{x^2+\frac{1}{4}}=\frac{\left(x+\frac{1}{2}\right)^2}{x^2+\frac{1}{4}}-1\ge-1\)dấu "=" xảy ra <=>\(x+\frac{1}{2}=0\) <=>x=\(\frac{-1}{2}\) Vậy min=-1 khi x=-1/2