Question

图上这题怎么做？

Answer 1

图上这题怎么做？
具体解答
如图所示

Answer 2

f(x)=x^2+ax+3-a

f'(x)= 2x+a

f'(x)=0

x=-a/2

f''(x) = 2 >0 ( min)

case 1: a≤-2

f(-2) =4-2a +3 -a =7-3a

f(-2) ≥0

7-3a ≥0

a≤ 7/3

solution for case 1: a≤-2

case 2: -2<a<2

f(a) =2a^2 -a +3 ≥0

a≤-1 or a≥3/2

solution for case 2:  -2<a≤-1

case 3: a≥2

f(2)= 4+2a+3-a = 7+a

f(2) ≥0

7+a ≥ 0

a≥ -7

solution for case 2: a≥2

f(x)=x^2+ax+3-a  ≥0                  ,x∈[-2,2]

=>

case 1 or case 2 or case 3

a≤-2 or -2<a≤-1 or  a≥2

a≤-1 or a≥2