设f(x)满足af(x)+b(1/x)=c/x,其中a,b,c都是常数,且|a|≠|b|,①证明f(x)为奇函数②求f'(x)和 f''(x)

(1)af(x)+b(1/x)=c/x (1)af(-x) -b(1/x) = -c/x (2)(1)+(2)a(f(x)+f(-x))=0f(x) = -f(-x)f(x)为奇函数(2)af(x)+b(1/x)=c/xaf'(x)-b(1/x^2)=-c/x^2f'(x) = (b-c)/(ax^2)f''(x) = -3(b-c)/(ax^3)不好意思。。。打错了,应该是af(x)+bf(1/x)=c/x