Does anyone know a complete set of compulsory explanation videos about senior one math teachers, such as definition domain and value domain?

2.① The domain is negative infinity to 0, and 0 to positive infinity. Derivation =2+ 1+ 1/X? If the constant is greater than zero, the pitch will increase in the domain, ranging from negative infinity to positive infinity.

② Let t=X? , the original formula = 1/t+ 1, t is greater than or equal to 0, and the reference curve y = 1/x. When t=0, the maximum value is 1, so the range of values is (0,1). 4. Solution, the original formula = 2x-13/2-√ (13-4x)+3.5, let t=√ 13-4X, then the original formula =-t? /2-t+3.5, where t is greater than or equal to zero, and -b/2a=- 1, monotonically increasing on (0, positive infinity), that is, the range of values is (3.5, positive infinity). 5. ① Solve, let t=X-2, then the original formula = | t |+t+3 |, and then | t |+t+3 | is greater than or equal to | t |+3-t | = 3, then the value range is (3, positive infinity).

(2) solution, -b/2a= 1, the minimum value is -4 at 1, and the maximum value is 0 at X=- 1, so the range of values is (-4,0) 6. When a=0, y= 1, which satisfies the condition.

When a is not equal to 0, -b/2a=- 1, and to get the maximum value, a is less than 0 (the curve opens downward), and the maximum value is 1-a,

Is this the condition of this question? If so, it seems that we can't find a specific value, but this is the idea. Please check the problem again.

I hope I can help you.