Question

Xét tính chẵn lẻ của hàm số: f(x) = \(\hept{\begin{cases}x^3+1;x\le-1\\0;-1< x< 1\\x^3-1;x\ge1\end{cases}}\)

Answer 1

Với giá trị \(x_0\) bất kì:

- Nếu \(-1< x_0< 1\Rightarrow-1< -x_0< 1\)

\(\Rightarrow f\left(-x_0\right)=0=-0=-f\left(x_0\right)\)

- Nếu \(x_0\le-1\Rightarrow f\left(x_0\right)=x_0^3+1\)

\(x_0\le-1\Rightarrow-x_0\ge1\Rightarrow f\left(-x_0\right)=\left(-x_0\right)^3-1=-\left(x^3_0+1\right)=-f\left(x_0\right)\)

- Nếu \(x_0\ge1\Rightarrow-x_0\le-1\)

\(f\left(x_0\right)=x_0^3-1\)

\(f\left(-x_0\right)=\left(-x_0\right)^3+1=-\left(x_0^3-1\right)=-f\left(x_0\right)\)

Vậy \(f\left(-x_0\right)=-f\left(x_0\right)\) \(\forall x_0\in R\Rightarrow f\left(x\right)\) là hàm lẻ