Question

①求+y=(x^2-5x+2)/(x+12)最小值

Answer 1

基本不等式的应用中，强调“何时取等号?”。

详情如图所示:

供参考，请笑纳。

也是考察对勾函数的单调性。

Answer 2

最小化 | (x^2 + 5 x + 2)/(x + 12)

min{(x^2 + 5 x + 2)/(x + 12)} = 2 sqrt(86) - 19 at x = sqrt(86) - 12

Answer 3

y = (x^2-5x+2)/(x+12)
y' = [(2x-5)(x+12)-(x^2-5x+2)]/(x+12)^2
= (x^2+24x-62)/(x+12)^2, 得唯一正驻点 x = √206-12 ≈ 2.3527
，最小值 y = [(√206-12)^2-5(√206-12)+2]/√206
= [(√206-12)^2-5(√206-12)+2]/√206 = 2√206-29 ≈ - 0.2946

Answer 4

y=(x^2+5x+2)/(x+12)
=(x^2+24x+144-19x-228+86)/(x+12)
=[(x+12)^2-19(x+12)+86]/(x+12)
=(x+12)+86/(x+12)-19
因为x>=0，所以x+12>=12>√86
所以y=(x+12)+86/(x+12)-19
>=12+86/12-19
=1/6
即最小值为1/6