f'(x)=2^x+x*2^xln2=(1+xln2)*2^x=0,则x=-1/ln2.
f(x)在区间(-无穷,-1/ln2)上递减,在区间(-1/ln2,+无穷)上递增.
所以,x=-1/ln2是f(x)的极小值点,即x0=-1/ln2.
啊?那个f'(x)怎么求的啊尤其是2^x的导数??thank
2^x=e^(ln2^x)=e^(xln2)(2^x)'=[e^(xln2)]'=[e^(xln2)]ln2=[e^(ln2^x)]ln2=2^xln2。记住:(a^x)'=a^xlna,这是公式。