已知f(x)在定义域(0,+无穷)为减函数f(xy)=f(x)+f(y),f(1/2)=1 (1)求f(1) (2)解不等式f(-x)+f(3-x)>=-2
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![已知f(x)在定义域(0,+无穷)为减函数f(xy)=f(x)+f(y),f(1/2)=1 (1)求f(1) (2)解不等式f(-x)+f(3-x)>=-2](/uploads/image/z/1120029-69-9.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%E5%9C%A8%E5%AE%9A%E4%B9%89%E5%9F%9F%280%2C%2B%E6%97%A0%E7%A9%B7%29%E4%B8%BA%E5%87%8F%E5%87%BD%E6%95%B0f%28xy%29%3Df%28x%29%2Bf%28y%29%2Cf%281%2F2%29%3D1+%EF%BC%881%EF%BC%89%E6%B1%82f%281%29+%282%29%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28-x%29%2Bf%283-x%29%3E%3D-2)
已知f(x)在定义域(0,+无穷)为减函数f(xy)=f(x)+f(y),f(1/2)=1 (1)求f(1) (2)解不等式f(-x)+f(3-x)>=-2
已知f(x)在定义域(0,+无穷)为减函数f(xy)=f(x)+f(y),f(1/2)=1 (1)求f(1) (2)解不等式f(-x)+f(3-x)>=-2
已知f(x)在定义域(0,+无穷)为减函数f(xy)=f(x)+f(y),f(1/2)=1 (1)求f(1) (2)解不等式f(-x)+f(3-x)>=-2
1.因为f(xy)=f(x)+f(y)恒成立,所以令x=1,y=1/2,可得f(1/2)=f(1)+f(1/2),解得,f(1)=0
2.思想是利用减函数的条件脱去f,方可解出x
因为f(xy)=f(x)+f(y),所以不等式变形为
f[x(x-3)]>= -2.
因为f(1/2)=1,所以右端的-2看成-2*1,继续变形为
f[x(x-3)]>= -2f(1/2);
f[x(x-3)] >= -[f(1/2) + f(1/2)]
右端再利用f(xy)=f(x)+f(y)合并,得
f[x(x-3)] >= -f(1/4);
f[x(x-3)] + f(1/4) >= 0;
同理,左端再利用公式合并
f{1/4[x(x-3)]} >= 0;
此时利用第一问求出的f(1)=0替换右端的0,得
f{1/4[x(x-3)]} >= f(1);
因为是减函数,所以
1/4[x(x-3)] <= 1;
化简得(x-4)(x+1) <= 0;
解得-1<= x <= 4;
另外定义域是正数,所以1/4[x(x-3)] > 0;
解得x>3 或 x<0
综合两个结果,最后的答案是-1 <= x < 0 或 3 < x <= 4;