設函數f(x)=sinx,g(x)=-9(x/π)^2+9(x/π)-3/4,則使g(x)≥f(x)的x的取值范圍是?請寫明詳細過程,謝謝!
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解:g(x)=-9(x/π)^+9(x/π)-3/4=-9(x/π-1/2)^+3/2∴f(x),g(x)的對稱軸都是x=π/2,在x∈[0,π]上開口向下的,當x=π/6或x=5π/6時,f(x)=g(x)=1/2.則使g(x)≥f(x)的x的取值范圍,x∈(π/6x,5π/6).
設函數f(x)=sinx,g(x)=-9(x/π)^2+9(x/π)-3/4,則使g(x)≥f(x)的x的取值范圍是?請寫明詳細過程,謝謝!
解:g(x)=-9(x/π)^+9(x/π)-3/4=-9(x/π-1/2)^+3/2∴f(x),g(x)的對稱軸都是x=π/2,在x∈[0,π]上開口向下的,當x=π/6或x=5π/6時,f(x)=g(x)=1/2.則使g(x)≥f(x)的x的取值范圍,x∈(π/6x,5π/6).