Monotonicity and maximum small values of math functions

Solution: 1, let x=y=1

f(1)=f(1)+f(1)，f(1)=02、x>0 x+2>0

f(x)+f(x+2)=f(x 2+2x) 2f(1 9)=f(1 3)+f(1 3)=2 because the function y=f(x) is a subtractive function defined on r+.

f(x)+f(x+2)=f(x^2+2x)＜2=f(1/9)x^2+2x>1/9

x> 10 3-1 or x<-1-10 3

Let x=y=1 f(xy)=f(x)+f(y), and bring in f(1)=f(1)+f(1).

f(1)=0

f(1 9) = f(1 3) + f(1 3) = 2f(x) + f(x+2) = f(x(x+2)) because f(x) + f(x+2) 2

So f(x(x(x+2))1 9

x> 10 3-1 or x<-1-10 3

Because x>0.

x>√10/3-1

Solution:1Let x=y=1, then there is f(1)=f(1)+f(1), so f(1)=0.

2.Let x=1 3 and y=3, then there is f(1)= f(1 3)+f(3), that is, 0=1+f(3), f(3)=-1

Then let x=3, y=1 9, then there is f(1 3)=f(3)+f(1 9), and f(1 9)=2

And because the inequality f(x)+f(x+2)<2 is equivalent to.

f(x(x+2))<2, because f(x) is a subtraction function, and f(1 9)=2 then the problem is finally reduced to x(x+2)>1 9, and this inequality can be solved.

I agree with the opinion of the first floor.

1. y=-x + root number x=-(x-1 2 root number x) squared + 1 4, when x=1 4 is the smallest, is 1 Because 1-x squared is definitely "0", it is obtained that -1" or equal to x" or equal to 1, when x is in [-1,0), the function is an increasing function, and when x is in (0,1], it is a subtraction function. 3. The numerator and denominator are multiplied by the root number (x square + 1).

x, the molecule is the only one after the same rationalization, the original formula = 1 root number (x square pure lead + 1).

x, the denominator is the root number (x squared + 1) + x, when x is in [0, + infinite) is a good number for the increase function, and the original formula is definitely greater than 0, so the original formula is a subtraction function when it is [0, + infinite), so when x = 0 is the maximum, what is it? 1,y=1 x+x, since both x and 1 x are positive at (0,+infinity), we get 1 x+x> or equal to 2*x*1 x=2 if and only if x=1 x, i.e., when x=1. 2. If y=1+x x=1+1 x, because 1 x is a subtraction function at [0,+infinity), then when x=infinity, y is the smallest, which is 1

All four questions are simple! Find the first derivative. Then let the derivative and other rents hide in the zero cons hall, and come up with an x, which is an extreme value within the value range! The derivative is greater than zero and the derivative is less than zero monotonically decreasing.

There are many ways to do it

In the case of the quadratic function in junior high school, you can judge the monotonicity by looking at the quadratic coefficient and the axis of symmetry, and substitute the square shouting Qingcheng of the axis of symmetry into the quadratic function, which is the maximum value (or minimum value).

The quadratic function recipe is also fine, in the form of y=ax 2+bx+c.

If it's high school.

Use the method of derivation (which can be differentiated).

For unary functions.

When the first derivative is positive, it is monotonically increasing.

When the first derivative is negative, it is monotonically decreasing.

The first derivative = 0 is a necessary condition to get the maximum value (the second derivative is required for sufficient conditions) for the binary comb function.

Let's calculate the partial derivative (similar to a univariate function).

There is also the method of definition.

y=1/x+1/(1-x)=1/[x(1-x)]=1/[-x-1/2)²+1/4]

Let g(x)=-x-1 2) +1 4, symmetrically talk about taking the pivot pin front x=1 and lose 2

When x = 1 2, y has a minimum value of 4

y=1/x + 1/(1-x)

1/x(1-x)

1/(x-x^2)

When x=1 2, the orange takes x-x 2=1 4 as the maximum, so it is simple.

The minimum value of the function is 4.

There are many ways to do it

In the case of the quadratic function in junior high school, the monotonicity can be judged by looking at the quadratic term ruler coefficient and the axis of symmetry, and the equation of the axis of symmetry is substituted into the quadratic function, which is the maximum value (or minimum value).

The quadratic function recipe is also fine, in the form of y=ax 2+bx+c.

If it's high school.

Use the method of derivation (which can be differentiated).

For unary functions.

When the first derivative is positive, it is monotonically increasing.

When the first derivative is negative, it is monotonically decreasing.

The first derivative = 0 is necessary to get the maximum value (the second derivative is required for sufficient conditions) for binary functions.

Let's calculate the partial derivative (similar to a univariate function).

There is also the method of definition.

That is, in the defined field, arbitrarily take x1, x2 (let x1

That is, for any x1, x2 belongs to the function definition domain (x1 x2), and there are f(x1)f(x2), then the function increases (decreases) monotonically

Here are the definitions.

The monotonicity of a function can also be called the addition or decrease of a function. When the independent variable of the function f(x) increases (or decreases) within its defined interval, and the value of the function f(x) also increases (or decreases), the function is said to be monotonionic in that interval.

The maximum value of the function.

Let the domain of the function y=f(x) be r, and if x0 r exists, the same is true for any x r, and x≠x0, there is a minimum value of f(x).

1.Let x=y=1, then f(1*1)=f(1)+f(1), and f(1)=0

2.Let x=y=1 3, then f(1 3)+f(1 3)=f(1 9)=2 let x=x, y=x+2, then f(x)+f(x+2)=f(x 2+2*x) "Reverse 2

Because f(x) is monotonically decreasing over the defined domain, it gets.

x 2+2*x>1 9, so you can scatter and carry x<-(sqr(10) 3-1 or x><(sqr(10) 3-1

1) In the identity f(xy)=f(x)+f(y), let x=y=1, f(1)=f(1)+f(1), f(1)=0;

2) In the identity f(xy)=f(x)+f(y), let x=y=1 3, f(1 9)=f(1 3)+f(1 3)=2f(1 3), f(1 3)=1, f(1 9)=2, the domain of the function is (0, in f(x)+f(x+2)<2, Lu Hebi x>0 and x+2>0, pat is x>0, from the identity, f(x)+f(x+2)=f(x(x+2)), f(x) is a subtractive function on (0, and the inequality f(x)+f(x+2)<2 can be reduced to.

f(x(x+2))

1 9, solution (3-10) 3