## Calculus: Early Transcendentals 8th Edition

When $x\ge0,$ $f(x)=x|x|=x\times x=x^{2}$ $f(-x)=-x|-x|=-x\times x=-x^{2}$ $f(-x)=-f(x)$ When $x\lt0,$ $f(x)=x|x|=x\times-x=-x^{2}$ $f(-x)=-x|-x|=-x\times-x=x^{2}$ $f(-x)=-f(x)$ $\therefore f(x)$ is even.