Question

已知函數(shù)f(x)和g(x)的圖象關(guān)于原點(diǎn)對(duì)稱,且f(x)=x²+2x.

(1)求函數(shù)g(x)的解析式；

(2)解不等式g(x)≥f(x)-|x-1|.

Answer 1

解析:(1)設(shè)函數(shù)y=f(x)的圖象上任一點(diǎn)Q(x₀,y₀)

關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為P(x,y),?

∵點(diǎn)Q(x₀,y₀)在函數(shù)y=f(x)的圖象上,

∴-y=x²-2x,即y=-x²+2x.?

∴g(x)=-x²+2x.?

(2)由g(x)≥f(x)-|x-1|可得2x²-|x-1|≤0.?

當(dāng)x≥1時(shí),2x²-x+1≤0,此時(shí)不等式無解.

當(dāng)x＜1時(shí),2x²+x-1≤0,得-1≤x≤

∴原不等式的解集為［-1,］.