Đáp án:
b. \(f\left( { - \frac{3}{2}} \right) = - \frac{{17}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
f\left( x \right) = 2x - {x^2} + 2\left| {x + 1} \right|\\
a.\left[ \begin{array}{l}
f\left( x \right) = 2x - {x^2} + 2\left( {x + 1} \right)\left( {x \ge - 1} \right)\\
f\left( x \right) = 2x - {x^2} - 2\left( {x + 1} \right)\left( {x < - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
f\left( x \right) = 2x - {x^2} + 2x + 2\\
f\left( x \right) = 2x - {x^2} - 2x - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
f\left( x \right) = - {x^2} + 4x + 2\\
f\left( x \right) = - {x^2} - 2
\end{array} \right.\\
b.Thay:x = - \frac{3}{2}\left( {x < - 1} \right)\\
\to f\left( x \right) = - {x^2} - 2\\
\to f\left( { - \frac{3}{2}} \right) = - {\left( { - \frac{3}{2}} \right)^2} - 2\\
= - \frac{{17}}{4}
\end{array}\)
