Question

1.y=x方+9分之x求最大值2.y=x4次方+9分之x方求最大值

Answer 1

(1)y＝x/(x²+9)

＝1/(x+9/x)

≤1/[2√(x·9/x)]

＝1/6.

∴x＝9/x→x＝3时,

y|max＝1/6.

(2)y＝x/(x^4+9)

→1/y＝x³+9/x

→1/y＝x³+3/x+3/x+3/x

≥4·(x³·3/x·3/x·3/x)^(1/4)

＝4×3^(3/4)

∴y≤1/[4×3^(3/4)],

即:y|max＝1/[4×3^(3/4)].

此时,x³＝3/x,x＝3^(1/4).