4．已知函数f(x)={a+ax,x⩾03+(a−1)x,x<0(a>0f(x)=\left\{{\left.\begin{array}{l}{a+{a^x},x\geqslant 0}\\ {3+({a-1})x,x<0}\end{array}\right.}\right.(a>0f(x)={a+ax,x⩾03+(a−1)x,x<0(a>0且a≠1)a\ne 1)a=1)，则"a⩾3a\geqslant 3a⩾3"是"f(x)f(x)f(x)在RRR上单调递增"的((( )))