High school math, ask for a check, do you do it right. Ask for a check. High School Mathematics

This conclusion is clearly problematic. Think of it this way (just draw it to verify):

For the function values f(a) and f(b) of the two endpoints of the function y=f(x) in the continuous closed interval [a,b], if f(a)*f(b)>0 is satisfied, then the function y=f(x) has an even number of zeros (including 0 zeros) in the open interval (a,b); If f(a)*f(b)<0 is satisfied, then the function y=f(x) has an odd number of zeros in the open interval (a,b); If f(a)*f(b)=0 is satisfied, then the function y=f(x) does not necessarily have zero points in the open interval (a, b), and the number of zeros when there is a zero point cannot be determined.

When y=0, there are at least two zeros in the interval (a, b).

When y<0, there is at least one zero in the interval (a, b).

When y>0, the interval (a, b) cannot determine whether there is a zero point, it is generally a quadratic cubic function, or you already know the function image, not the function image you are doing, it refers to the general shape of that kind, for example, all quadratic ones are a u or n, and cubic ones are s or inverted s-shaped.

**Don't be embarrassed to ask if you don't know.

This conclusion is obviously wrong, and the most important thing is to draw pictures. You look again, according to the first person.

The students did not find out the value range of the new variable t [-1, -1 2] after the commutation, which is the most important point to pay attention to in the commutation method. The logarithm with a base of 1 4 is a subtraction function, and when x belongs to 2 to 4, the minimum value is -1 at 4 t and the maximum value is -1 at 2 2The question in question 2 is the same as in question 1.

None of the new variables have been found to have a range of values.

(1) False, t [-1, -1 2].

So it's monotonically decreasing.

The value range is [23, 4, 7].

2) Not true. t∈[0,log(2,3)]

The value range is [0,2log(2,3)].

log(2,x）-log(1/2,x)=log(2,x)-[log(2,1/2)-log(2,x)]=2log(2,x)+1

log(2,x) is monotonically increasing, so the range is (1,2log(2,3)+1).

The train of thought is fine. It should be correct.

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