How to solve the functions of the math problem in the first year of high school

f(x)=2^[sinx]+3^[cosx]

x=0 f(x)=4 4>0+a→a<4

x (0, f(x)=2 f(x)>x+a 2>x+a a< 2 (x+a is an increasing function).

x=½πf(x)=3 3>0+a→a<3

x∈(½f(x)=4/3 f(x)>x+a→4/3>x+a→a<4/3-π

x∈(π3π/2] f(x)=5/6 f(x)>x+a→5/6>x+a→a<5/

x∈(3π/2,2π) f(x)=3/2 f(x)>x+a→3/2>x+a→a<3/2-2π

x=2π f(x)=4 4>2π+a→a<4-2π

min(4,½π2,3,4/3-π,5/,3/2-2π,4-2π)=3/2-2π

The inequality is constant and holds a<3 2-2

This student, first of all, the maximum value of f(x) is 4, and the minimum value is 0, so the maximum value of f(x1)-f(x2)丨 is 4, and then 2021 4=505 is more than 1, which means that n=505 cannot make the original formula = 2021, and 丨f(x1)-f(x2)丨+......丨f(xn-1)-f(xn)丨There is an n-1 group, so n-1=506,n=507 is how it comes, this question is not difficult but it tests whether the details of the students are careful, I hope it can help you!

The domain of the function tan can be limited to (-2, 2), so the domain of tan(x-v4) is (-2-v4, 2-v4), i.e. (-3 v4, -v4).

The tan() function cannot be equal to 2 in parentheses, and the function period is

i.e. x- 4≠k* + 2 (k is a natural number).

x≠k* +3 4 (k is a natural number).

Sorry it's inconvenient to write by hand, I'll type it, the answer should be, root number two x 4,,, first you look at the above definition of the domain, after substitution ( ) in the range is 1 2 2,, the following is also the same function, so ( ) in the value range is the same, so x, the value is ,, root number two 4, ( 2 - root number two 1 2,).

The definition of a domain is the set of all the values of the argument that satisfy the meaning of the function in the case of making the function meaningful, that is, the range of values of the argument variable. Let it make sense means that for example, the true number of the logarithmic function must be greater than zero, otherwise it is meaningless, and then the number in the quadratic root must be greater than or equal to zero, the denominator must not be zero, etc., and if it is not satisfied, it will make the function meaningless, and the first function of this problem, the independent variable is represented by the x square of 2, let it be less than or equal to one, greater than or equal to minus, and solve an x range, and the second function guarantees that x is greater than zero, and the two solution sets take the intersection because they must be satisfied at the same time, so take the intersection, and then you can draw a picture or calculate it directly, because the logarithmic function is constant or decreasing, Directly substitute the maximum and minimum values of the final x-range into the range to solve the value range. I forgot the number in the question, and I couldn't look back at this answer, and I was afraid that the words would be typed in vain, and then I explained the reason, and there was no problem in doing the question.

You've learned. f(x)=ax+b/x』

This function is called the checkmark (double tick) function, and the image is distributed in one or three quadrants. Words can't express it clearly!

f'(x)=3-2/x2

So when 2 x2<3, i.e., x2>2 3, then x> root number 6 of 3 or root number 6 of x<-3 points, since x is in the (-1, + infinite) interval, so when the root number of -13 is 6, f(x) is the increasing function; When the root of -3 is 6< x < root number 6 of the third, f(x) is a subtraction function.

Using the algebraic method of the interval "-1. +Infinity" We can casually bring in two trees greater than -1 and see the changes in f"x"!

f(9)=f(3*3)=f(3)+f(3)=1+1=2f(27)=f(3*9)=f(3)+f(9)=1+2=3f(x)+f(x-8)=f(x*(x-8))<2=f(9) multiplication function x(x-8)<9

x^2-8x-9<0

Left root x=9 x=-1

10 and x>0

So 8

(1) f(9)=f(3*3)=f(3)+f(3)=2f(27)=f(3*9)=f(3)+f(9)=3(2) because f(9)=2 and the function is an increasing function in the defined domain, and because f(x 2-8x) = f(x)+f(x-8), 00,x-8>0

8

f（9）=f（3*3）=f（3）+f（3）=2

In the same way, f(27)=f(9*3)=2+1=3

The domain of f(x) is a non-negative real number, which increases monotonically on the domain of definition, and it is known that f(9)=2, we know that f(0)=f(0*0)=2f(0), so f(0)=0Then f(x)+f(x-8)=f(x's squared -8x)<2, so 0<=x's squared -8x<9, then x>=8 and -1< x<9, so 8<=x<9

1) f(9) = f(3*3) = f(3)+f(3) = 1+1 = 2

Similarly: f(27) = f(9)+f(3) = 2+1 = 32)f(x)+f(x-8) = f( x(x-8 ) 2=f(9).

f(x) is the increment function, and the domain is defined as: x(x-8) <9: x>0, x-8>0

So 8