Question 14: The number of extreme points of the function (y = x + sqrt {2{x^2} + 1} ) is

Specified Set \(D=\mathbb{R}\)

\(y’ = 1 + \frac{{2x}}{{\sqrt {2{x^2} + 1} }} = \frac{{\sqrt {2{x^2} + 1} + 2x} }{{\sqrt {2{x^2} + 1} }}\)

\(y’ = 0 \Leftrightarrow \sqrt {2{x^2} + 1} = – 2x\)

\( \Leftrightarrow \left\{ \begin{array}{l}x \le 0\\2{x^2} + 1 = 4{x^2}\end{array} \right.\)

\( \Leftrightarrow \left\{\begin{array}{l}x \le 0\\\left[\begin{array}{l}x=–\frac{{\sqrt2}}{2}\left({face}\right)\\x=\frac{{\sqrt2}}{2}\left({type}\right)\end{array}\right\end{array}\right\)[\begin{array}{l}x=–\frac{{\sqrt2}}{2}\left({nhan}\right)\x=\frac{{\sqrt2}}{2}\left({loai}\right)\end{array}\right\end{array}\right\)

Variation table

So the given function has 1 extreme.