The root of the equation and the zero point of the function! 1. Find the zero point of the function ①y=-x^2+x+6 ②y=(x^2-2)(x^2-3x+2) 2. The interval of the root of square root lgx + x = 0 is () A.(-∞,0) B.(0,1) C.(1,2) D.(2,4) 3. It is known that the image of the function y = f (x) is continuous, if there is a corresponding value table as follows X 1 2 3 4 5 6 y 123.56 21.45 -7.82 11.45 -53.76 -128.88 Then the function y = f (x) has at least () A. 2 B.3 C.4 D 5 A function with zeros in the interval [3,5] is () A.f(x)=2xln(x-2)-3; B.f(x)=-x^3-3x+5 C.f(x)=2^x-4 D.f(x)=-1/x+2 4. If the two zeros of the function f (x) = x ^ 2 - (T-2) x + 5-T are greater than 2, then the value range of T is_____ 5. The quadratic equation x ^ 2 + (m-1) x + 1 = 0 has unique solution in the interval [0,2], Then the value range of the real number M_________ 6. Let f (x) = ax + 2A + 1 (a ≠ 0)

The root of the equation and the zero point of the function! 1. Find the zero point of the function ①y=-x^2+x+6 ②y=(x^2-2)(x^2-3x+2) 2. The interval of the root of square root lgx + x = 0 is () A.(-∞,0) B.(0,1) C.(1,2) D.(2,4) 3. It is known that the image of the function y = f (x) is continuous, if there is a corresponding value table as follows X 1 2 3 4 5 6 y 123.56 21.45 -7.82 11.45 -53.76 -128.88 Then the function y = f (x) has at least () A. 2 B.3 C.4 D 5 A function with zeros in the interval [3,5] is () A.f(x)=2xln(x-2)-3; B.f(x)=-x^3-3x+5 C.f(x)=2^x-4 D.f(x)=-1/x+2 4. If the two zeros of the function f (x) = x ^ 2 - (T-2) x + 5-T are greater than 2, then the value range of T is_____ 5. The quadratic equation x ^ 2 + (m-1) x + 1 = 0 has unique solution in the interval [0,2], Then the value range of the real number M_________ 6. Let f (x) = ax + 2A + 1 (a ≠ 0)

Today is your test for me
1.(1).(X-3)(X+2)=0,X1=3,X2=-2.
(2).X^2-2=0,(X^2-3X+2)=0,(X-2)(X-1)=0.
X1=√2,X2=-√2,X3=2,X4=1.
2. The interval of the root of square root lgx + x = 0 is (b)
10> 0, a is impossible, lgx + X ≥ 0. Lgx must be negative root to be zero. In C and D, lgx is positive root, only B has negative root
3. Then the zero point of function y = f (x) in interval [1,6] has at least (b)
It can be seen from the rough drawing,
A function with zeros in interval [3,5] is (a)
Substitute x = 3 and x = 5 into each equation to see if there are two zeros, only a
4. If the two zeros of function f (x) = x ^ 2 - (T-2) x + 5-T are greater than 2, then the value range of T is t > 6
Axis of symmetry x = (T-2) / 2 > 2, t > 6
5. The quadratic equation x ^ 2 + (m-1) x + 1 = 0 has unique solution in the interval [0,2],
Then the value range of the real number M____ m