Cho hàm số \(y=f\left( x \right)\) liên tục trên \(\mathbb{R}\), có đồ thị \({f}'\left( x \right)\) như hình vẽ.

Chọn A

Ta có \(g\left( x \right)=f\left( -{{x}^{2}}+x \right)\)\( \Rightarrow {g}'\left( x \right)=\left( -2x+1 \right){f}'\left( -{{x}^{2}}+x \right)\).

\(\Rightarrow {g}'\left( x \right)=0\)\( \Leftrightarrow \left( -2x+1 \right){f}'\left( -{{x}^{2}}+x \right)=0\)

\(\Leftrightarrow \left[ \begin{align} & -2x+1=0 \\ & {f}'\left( -{{x}^{2}}+x \right)=0 \\ \end{align} \right.\)

\(\Leftrightarrow \left[ \begin{align} & x=\frac{1}{2} \\ & -{{x}^{2}}+x=0 \\ & -{{x}^{2}}+x=2 \\ \end{align} \right.\)

\(\Leftrightarrow \left[ \begin{align} & x=\frac{1}{2} \\ & x=1 \\ & x=0 \\ \end{align} \right.\)

Do đó \({g}'\left( x \right)>0\)\( \Leftrightarrow \left( -2x+1 \right){f}'\left( -{{x}^{2}}+x \right)>0\)

\(\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} - 2x + 1 > 0\\ f'\left( { - {x^2} + x} \right) > 0 \end{array} \right.\\ \left\{ \begin{array}{l} - 2x + 1 < 0\\ f'\left( { - {x^2} + x} \right) < 0 \end{array} \right. \end{array} \right.\)
\(\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x < \frac{1}{2}\\ \left[ \begin{array}{l} - {x^2} + x > 2\\ - {x^2} + x < 0 \end{array} \right. \end{array} \right.\\ \left\{ \begin{array}{l} x > \frac{1}{2}\\ 0 < - {x^2} + x < 2 \end{array} \right. \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x < \frac{1}{2}\\ \left[ \begin{array}{l} x > 1\\ x < 0 \end{array} \right. \end{array} \right.\\ \left\{ \begin{array}{l} x > \frac{1}{2}\\ 0 \end{array} \right. \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l} x < 0\\ \frac{1}{2} \end{array} \right.\)

Bảng biến thiên

Vậy hàm số có 1 điểm cực tiểu.