If the derivative of function f (x) is f ˊ (x) = - SiNx, then the inclination angle of tangent line at point (π / 3, f (π / 3) is

If the derivative of function f (x) is f ˊ (x) = - SiNx, then the inclination angle of tangent line at point (π / 3, f (π / 3) is

fˊ(x)=-sinx
tanA=fˊ(π/3)=-sinπ/3=-√3/2
The angle of inclination of tangent at point (π / 3, f (π / 3) is as follows:
π-arctan√3/2

Let f (x) be differentiable and f (0) = 0. Let f (x) = f (x) (1 + | SiNx |), and find f '(0)

When x > 0, there is f (x) = f (x) (1 + SiNx)
When x

The image of function y = 2Sin (x - π / 6) is shifted π / 6 unit length to obtain the image of function y = 2Sin (x - π / 3) Explain why

The length of the (2x) sin / π - function is shifted to the right, and the image length is obtained
Because in rectangular coordinate system, the translation of geometry conforms to the law of left plus right minus, so the right translation is X - π / 6 - π / 6 = x - π / 3, so it is right

The general procedure of drawing function image by point tracing method is___ 、 ___ 、 ___ .

The general steps of drawing function images with point tracing method are listing, tracing points and connecting lines,
So the answer is: list, point, line

How to draw the image of y = x + 3 function

This is a linear function, and its image is a straight line
You can use the simplest point tracing method, that is, two points determine a straight line
You can take two values, x = 1 and x = 0, and find the corresponding y value respectively. There are two points
（0,3）,（1,4）
The two points are projected on the coordinate axis and connected into a straight line

A mathematical problem about function and function image In this paper, the definition of two straight lines determined by the images of two first-order functions is given: let the image of the first-order function y = HX + B (H ≠ 0) be a line L, and the image of a function of degree y = ex + n (E ≠ 0) be a straight line S. if h = E and B ≠ n, we call the line L parallel to the line s The function expression of the straight line passing through point P (1,4) and parallel to the known line y = - 2x-1 and the points (2 points) that the image passes through Thank oil

Let the straight line be y = KX + B
Then by the definition of parallelism, k = - 2 and B ≠ - 1
Because the line passes through P, 4 = - 2 * 1 + B, B = 6
So the straight line is y = - 2x + 6
Just take two points on this line, such as P and (0, 6)

Given the function f (x) = 2lnx-x ^ 2-ax (1), find the monotone interval of the function (2) If the function f (x) has two different zeros x1, X2 and x1 Math homework for users on October 31, 2017 report Use this app to check the operation efficiently and accurately!

(1) (2) let the midpoint of X1 and X2 be x0, because px0f '(x) = 2 / x-2x-a, the second derivative of F (x) is monotonically decreasing, so f' (PX1 + qx2) = 0 is known as (1) 2lnx1-x1 ^ 2-ax1 = 0 (2) 2lnx2-x2 ^ 2-ax

It is proved that the function f (x) = 3x + 2 is an increasing function on (- ∞, + ∞) Quick

Take any X1 < x2
F(x2) - F(x1)
= (3x2 + 2) - (3x1 + 2)
= 3x2 - 3x1
= 3(x2 - x1) > 0
So f (x2) > F (x1)
So f (x) is an increasing function on (- ∞, + ∞)

Find the definition domain and value domain of the function y = log2 ^ (x ^ 2 + 2x + 5), and find the specific process

Definition domain (- ∞, + ∞)
x²+2x+5 = (x+1)² +4 ≥ 4；
So log2 (x 2 + 2x + 5) ≥ log2 (4) = 2;
Range: [2, + ∞)

Given the function f (x) =sin2xcos2x- the root sign 3sin 2 2x, find the value range of F (x) on the interval [0, π /4]

f*x(=1/2*sin4x-√3(1-cos4x)/2
=sin4xcosπ/3+cos4xsinπ/3-√3/2
=sin(4x+π/3)-√3/2
π/3<=4x+π/3<=4π/3
So - √ 3 / 2 < = sin (4x + π / 3) < = 1
Subtract from 3 -
Range [- √ 3,1 - √ 3 / 2]