f(x)=cos(2arccosx)+4sin[arcsin(x/2)]
= 2[cos(arccosx)]^2 - 1 + 4 * (x/2)
= 2x^2 -1 + 2x
= 2x^2 + 2x - 1
= 2(x^2 + x) - 1
= 2(x^2 + x + 1/4 - 1/4) - 1
= 2[(x + 1/2)^2 - 1/4] - 1
= 2(x + 1/2)^2 - 3/2
x 的定義域為 [-1,1]
因此
當(dāng) x = -1/2 時,f(x)取最小值 -3/2
當(dāng) x = 1 時,f(x) 取最大值 3