How to find the maximum and minimum value of the higher function!!

f(x)=√x/(x+1)

You can find 1 f(x)=(x+1) x

minimum value. 1/f(x)

x1 x because x 0, so set.

t = x thus.

1/f(x)=t²

1/t²=(t1/t)²

While. t-1 t=0, i.e. t=1, 1 f(x) has a minimum value of 2 thereby. While.

When x=1, f(x) has a maximum value of 1 2.

The commutation method is available.

Let t = root number x, then t>=0, x=t 2

f(x)=t/(t^2+1)

From the mean inequality, t 2 + 1 > = 2t, so there is f(x)< = 1 2, the maximum value is 1 2, when t = 1, that is, x = 1 to take the maximum value.

Swap the element, change the root number x to a, divide a at the same time up and down, and the following becomes a tick function, and the top is one, you should know it. The most important thing for this type of question is to define the domain.

When a = -1, the function f(x) = x -2x+2 axis of symmetry is 1 because this is a parabola with an opening upward, so the minimum value at the axis of symmetry is 1 because -5 is far away from the axis of symmetry, so the maximum value is obtained at -5, and the maximum value is 37, what else is there to understand.

The maximum value of the function is solved.

1. Observation method: For simple functions, the maximum value can be directly calculated after the known analytical formula is appropriately deformed.

2. Discriminant method: Some functions can be organized into quadratic forms about fx after appropriate deformation Since is a real number, such functions can be found with discriminant expressions. However, it is necessary to remove (or retrieve) the part of the function value range that expands (or shrinks) during the deformation

3. Monotonicity method: If the function is bounded in each monotonic interval within the defined domain (there may be only an upper bound without a lower bound or only a lower bound without an upper bound), the value range on each interval can be found first, and then the value range of the original function can be determined by their union, so as to obtain the maximum value of the function.

Fourth, the mean inequality method: if, , when it is a fixed value, then if and only if =, there is a minimum value; If it is a fixed value, then there is a maximum value if and only if =.

5. Trigonometric substitution: For the maximum value of some functions, trigonometric substitution can be used to solve it cleverly. In the case of substitution, the corresponding substitution can be made according to the analytical formula of different functions. Such as: may be ordered; Can be ordered () can be ordered, etc.

6. Combination of numbers and shapes: It is also a common method to solve the problem of maximum value by giving some abstract analytical formulas to geometric meanings, and then converting the information between "number" and "shape" through the attributes and quantity relations of figures.

7. Clever coordinate method: For the solution of the maximum value of irrational functions, the position of some special points in the Cartesian coordinate system can be used to solve the problem. 8. Use the modulus of complex numbers:

It is also an effective way to solve the maximum value of some irrational functions by treating irrational numbers as complex modulos, and then using the concept of complex modulus and the inequality of complex modulus. However, it should be noted that the modulo of the sum of all complex numbers must be satisfied.

In general, the maximum value of a function is divided into the minimum value of the function and the maximum value of the function. In simple terms, the minimum value is the minimum value of the function value in the defined domain, and the maximum value is the maximum value of the function value in the defined domain. The geometric significance of the maximum (small) value of the function - the ordinate of the highest (low) point of the function image is the maximum (small) value of the function.

1. Matching method: the function of the form, according to the extreme point of the quadratic function or the value of the boundary point to determine the maximum value of the function.

2. Discriminant method: a fractional function of the form, which is reduced to a quadratic equation with a coefficient of y about x. Since, 0, to find the maximum value of y, this method is easy to produce root increase, so it is necessary to test whether there is a solution to the corresponding x value when obtaining the maximum value.

3. Use the monotonicity of the function: first clarify the definition domain and monotonicity of the function, and then find the maximum value.

4. Use the mean inequality, the function of the form, and pay attention to the application conditions of positive, definite, etc., that is: a, b are positive numbers, is a fixed value, and whether the equal sign of a=b is true.

5. Commutation method: the function of the form, let and inversely solve x, substitute the above equation to obtain the function about t, pay attention to the definition range of t, and then find the maximum value of the function about t. There is also trigonometric commutation method, parameter commutation method.

6. Combination of numbers and shapes: For example, the left side of the formula is regarded as a function and the right side is regarded as a function, and their images are made in the same coordinate system, and their position relationships are observed, and the maximum value is obtained by using analytic geometry knowledge. Find the maximum value of the shape using the slope formula of the straight line.

7. Use the derivative to find the maximum value of the function: firstly, it is required to define the symmetry of the domain with respect to the origin and then judge the relationship between f(x) and f(-x): if f(x)=f(-x), even function; If f(x)=-f(-x), odd function.

For functions f:a->r, if AEA exists, so that for all XEAs, there is fix) In order to find the maximum and minimum values, the basic method is to first determine their existence, and then compare the functions at the stationary point, and define the value of the function at the endpoint of the domain or boundary point, the non-differentiable point, where the largest (small) is the maximum (small) value.

In many application problems, the existence of maximum and minimum values can often be determined by the context of the specific problem.

The first to use differential calculus to find the maximum and minimum values was Fermat. He discovered the extreme requisite called Fermat's theorem (not in its present form) and determined that the function reaches a maximum or minimum value at a stationary point. Extreme value problems have always been a concern for mathematicians, and there are several mathematical disciplines that study more complex extreme value problems, such as convex analysis, mathematical programming, variationalism, etc.

f(x)=√x/(x+1)

You can find 1 f(x)=(x+1) x

minimum value. 1/f(x)

x1 x because x 0, so set.

t = x thus.

1/f(x)=t²

1/t²=(t1/t)²

While. t-1 t=0, i.e. t=1, 1 f(x) has a minimum value of 2 thereby. While.

When x=1, f(x) has a maximum value of 1 2.

The commutation method is available.

Let t = root number x, then t>=0, x=t 2

f(x)=t/(t^2+1)

by mean inequalities.

t 2 + 1 > = 2t, so there is f(x) < = 1 2, the maximum value is 1 2, when t = 1, that is, x = 1 to get the maximum value.

When a = -1, the function f(x) = x -2x+2 axis of symmetry is 1 because this is a parabola with an opening upward, so the minimum value at the axis of symmetry is 1 because -5 is far away from the axis of symmetry, so the maximum value is obtained at -5, and the maximum value is 37, what else is there to understand.

(1)y=2x 2+4x+3=2(x+1) 2+1 when x=-1 has a minimum value of 1;

or y'=4x+4, when x=-1, y'=0;Functions have minimals;

2) y=-2x 2+3x-1=-2(x-3 4) 2+9 8-1 when x=3 4 has a maximum value of 1 8;

3) y=3x 2+5x+2=3(x+5 6) 2-25 12+2 when x=-5 6 has a minimum value of -1 12;

2(x+1) 2+1 with a minimum value of 1 and a maximum value of positive infinity.

2(x-3 4) 2+1 8, with a maximum value of 1 8 and a minimum value of negative infinity.

x+1) (3x+2), the minimum value is 5 4, and the maximum value is negative infinity.

recipes, or substitution vertex formulas.

1) When x=—1, the minimum value of y = 1

2) When x = 3 4, y max = 1 8

3) When x = —5 6, the minimum value of y = —1 12

Let m-n=t>0, then m=n+t

Original formula = (n+t) 2+1 nt>=4nt+1 nt (if and only if n=t).

2 root number (4nt*1 nt)=4 (if and only if nt=1 2).