Question

Cho hàm số f(x)\(=x+\sqrt{x^2+1}\). Tập giá trị của x để \(2x.f'\left(x\right)-f\left(x\right)\ge1\)

Answer 1

\(f'\left(x\right)=1+\dfrac{x}{\sqrt{x^2+1}}=\dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\)

\(2x.\dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}-\left(x+\sqrt{x^2+1}\right)\ge1\)

\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(\dfrac{2x}{\sqrt{x^2+1}}-1\right)\ge1\)

\(\Leftrightarrow\dfrac{2x}{\sqrt{x^2+1}}-1\ge\sqrt{x^2+1}-x\)

\(\Leftrightarrow2x-\sqrt{x^2+1}\ge x^2+1-x\sqrt{x^2+1}\)

\(\Leftrightarrow\left(x-1\right)^2-\sqrt{x^2+1}\left(x-1\right)\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1-\sqrt{x^2+1}\right)\le0\)

\(\Leftrightarrow x-1\ge0\) (do \(x-1-\sqrt{x^2+1}< 0;\forall x\))

\(\Leftrightarrow x\ge1\)