Find the monotonic interval and extreme value of the function y x 1 e arctanx

The monotonic increment interval of the function is: (-inf,-1] [0,inf); The monotonic reduction interval is: [-1,0].

Minimum: -e arctan(0)=-e(0)=-1

y'=e arctanx+(x-1)e (arctanx) (1+x 2)=e arctanx((x 2+x) (x 2+1)), which defines the domain.

Yes r. When 10, so the monotonic increase interval of the function is: (-inf,-1] 0,inf); The monotonic reduction interval is: [-1,0].

f in the left side of the domain of x=-1'(x) >0, f in the right side of x=-1'(x)

So the function takes a minimum at x=0: e arctan(0)=-e (0)=-1

Explanation of terms: Functions are related to inequalities and equations (elementary functions.

Let the value of the function be equal to zero, which corresponds to the independent variable from a geometric point of view.

is the abscissa of the intersection of the image and the x-axis; From an algebraic point of view, the corresponding independent variable is the solution of the equation. Also, replace the " " in the expression of the function (except for functions without expression) with " " or " " and replace the "y" with another algebraic expression.

The function becomes an inequality and the range of the independent variables can be found.

The set of input values x is called the defined domain of f; The set of possible output values y is called the range of f.

The domain of a function is the set of actual output values obtained by mapping f to all elements in the defined domain. Note that it is incorrect to call the corresponding domain a value range, and the value range of a function is a subset of the corresponding domain of the function.

Computer science.

, the data type of the parameter and the return value.

The defined domain and the corresponding domain of the subroutine are determined, respectively. Therefore, defining the domain and the corresponding domain is a mandatory constraint that is determined at the beginning of the function. On the other hand, the range is related to the actual implementation.

y'=e arctanx+(x-1)e (arctanx) (1+x 2)=e arctanx((x 2+x) (x 2+1)), the domain is r

When 10, so the monotonic increase interval of the function is: (-inf,-1] 0,inf); The monotonic reduction interval is: [-1,0].

f in the left side of the domain of x=-1'(x) >0, f in the right side of x=-1'(x)

So the function takes a minimum at x=0: e arctan(0)=-e (0)=-1

Introduction

Extremum is a basic concept of variational methods. The maximum or minimum value obtained by the functional within a certain range of the allowable function is called the maximum or minimum value, respectively, and is collectively referred to as the extreme value. The variable function that brings the functional to an extreme value is called an extreme function, and if it is a univariate function, it is often called an extreme curve.

Extremum is also known as relative extremum or local extremum.

Extremum is a collective term for "maximum" and "minimum". If the value of a function at a point is greater than or equal to the value of the function at any other point near that point, the value of the function at that point is said to be the "maximum" of the function. If the value of a function at a point is less than or equal to the value of the function at any other point near that point, the value of the function at that point is said to be the "minima" of the function.

y=(x-1)e^arctanx

y′=(x-1)′e^arctanx+(x-1)[e^arctanx]′

e^arctanx+(x-1)e^arctanx• [arctanx]′

e^arctanx+(x-1)e^arctanx• [1/(1+x^2)]

e^arctanx(1+(x-1)/(1+x^2))

Let y =0e arctanx 0 i.e. find 1+(x-1) (1+x 2)=(1+x 2+x-1) (1+x 2)=(x 2+x) (1+x 2)=0

Dock point x1=

When x -1x 0, y 0The function y is monotonically incrementing.

When -1 x 0, y 0The function y is monotonically decreasing.

When x = -1, y obtains a maximum value of -2e (arctan(-1))=-2e (- 4).

When x=0, y obtains a minimum value of -e (0)=-1

In summary. The monotonic increase interval of the function y=(x-1)e arctanx is x [-1] [0,

The monotonic reduction interval of the function y=(x-1)e arctanx is x [-1,0].

The maximum value of y is -2e (-4).

The minimum value of y is -1

y'=e arctanx+(x-1)e (arctanx) (1+x 2)=e arctanx((x 2+x) (x 2+1)), the domain is r

e arctanx>0, (x 2+1)>0, so y'=0, i.e., x 2+x, solution: x = 0 or -1

When 10, so the monotonic increase interval of the function is: (-inf,-1] 0,inf); The monotonic reduction interval is: [-1,0].

On the left side of x=-1, the mountain cherry trembles in the domain f'(x) >0, f in the right side of x=-1'(x)0

The argument is called as a function at x=0 to obtain a minima: -e arctan(0)=-e (0)=-1

y=(x-1)e^arctanx

y′=(x-1)′e^arctanx+(x-1)[e^arctanx]′

e^arctanx+(x-1)e^arctanx• [arctanx]′

e^arctanx+(x-1)e^arctanx• [1/(1+x^2)]

e^arctanx(1+(x-1)/(1+x^2))

Let y =0e arctanx 0 i.e. find 1+(x-1) (1+x 2)=(Senzhen 1+x 2+x-1) (1+x 2)=(x 2+x) (1+x 2)=0

Dock point x1=

When x, y 0The function y is monotonically incrementing.

When -1 x 0, y 0The function y is monotonically decreasing.

When x = -1, y obtains a maximum value of -2e (arctan(-1))=-2e (- 4).

When x=0, y obtains a minimum value of -e (0)=-1

In summary. The monotonic increase interval of the function y=(x-1)e arctanx is x [-this is bold, -1] [0,

The monotonic reduction interval of the function Hu Hui y=(x-1)e arctanx is x [-1,0].

The maximum value of y is -2e (-4).

The minimum of y is -1,2,

Answer]: B Examination Points] This question examines the monotony of the letter touching the shed and the number of acres laughing. [Test-taking guidance].

From the meaning of the title, y=x-e x +1,x r, spine lead pants y =1-e x, so y =0 to get x=0

So when x changes, y, y changes as follows:

x (-0) excitation 0 (0,+ f (x) +0 - f(x) monotonically increasing 0 monotonically decreasing The monotonic decreasing interval of the function is (0,+ The monotonically increasing interval is (- 0), and the function obtains a maximum at x=0, with a maximum of 0 and no minimum

y'=e x(x+2)+e x=e x*(x+3)let y'=0 because e x>0

So x+3=0

x=-3 because e x and x+3 are both manuscript finger increment functions.

The key is matched with y'is an increment function.

x = -3 when y'=0

So x0 so x-3, y is the increment function.

So x=-3 is the minimum random rotation point.

Minimum = e (-3) * (3 + 2).

1/e^3

Summary. Hello, glad to answer for you. The only monotonic interval of the function y=x(x-1) (1 3) is [1,+ the maximum point is x=1 and the extreme value is 0.

Finding the monotonic interval and extrema of the function y x(x 1) (1 3), hello, glad to answer for you. The only monotonic interval of the function limb collapse width y=x(x-1) (1 3) is [1,+ the maximum point is x=1, and the extreme value is 0

Can you elaborate on that a little bit more?

OK. You can ask for Y first'=1(x-1) (2 3)-2x(x-1) (5 3), y'=0, which means that x=1 is the maximum tolerance point; Seek y again''=6 (x-1) (5 3)+10x (x-1) (8 3), when x=1, y''> 0, it means that the function becomes convex at the point of the maximum value of filial piety, so x=1 is the maximum point of the function, and the extreme value is 0Hope it helps.