A brief and detailed process of transforming trigonometric function sin (X-Y) cosy + cos (X-Y) siny

A brief and detailed process of transforming trigonometric function sin (X-Y) cosy + cos (X-Y) siny

sin(x-y)cosy+cos(x-y)siny
=sin[(x-y)+y]
=sinx

If the function y = 3 + x2ln (1 + x1 − x), X ∈ [- 12, 12] is m, m respectively, then M + M=______ ．

Let g (x) = x2ln (1 + x1 − x), X ∈ [- 12, 12], then G (- x) = x2ln (1 − X1 + x) = - G (x), that is, G (x) is an odd function, ∵ g (x) max + G (x) min = 0, ∵ 3 + x2ln (1 + x1 − x), the maximum and minimum values of X ∈ [- 12, 12] are m, m, ∵ m + M = 6 respectively

The quadratic function f (x) satisfies f (4 + x) = f (- x), and f (2) = 1, f (0) = 3. If f (x) has minimum value 1 and maximum value 3 on [0, M], then the value range of real number m is______ ．

Let f (x) = a (X-2) 2 + H, ∵ f (2) = 1, f (0) = 3, ∵ H = 14a + H = 3, H = 1A = 12 ∵ f (x) = 12 (X-2) 2 + 1, let 12 (X-2) 2 + 1 = 3, then x = 0 or x = 4 ∵ f (x) has minimum value 1, maximum value 3 on [0, M], and the range of real number m is [2, 4]. So the answer is: [2, 4]

Let a linear function y = KX + B pass through point a (2, - 1) and point B, where point B is the intersection of the line - 2x + 1 and the y-axis

Point B is the intersection B (0,1) of the line y = - 2x + 1 and the Y axis
The point a (2, - 1) / b (0,1) satisfying the condition is on the line y = KX + B
2k+b=-1
b=1,k=-1
The expression of this function is y = - x + 1

Given that f (x) is a function defined on the interval [0, + ∞), and monotonically increasing on the interval, then the value range of X satisfying f (2x-1) < f (1 / 3) is

∵ f (- x) = f (x) ∵ f (x) is an even function
And ∵ f (x) increases monotonically in the interval [0, + ∞) and decreases monotonically in the interval (- ∞, 0)
To make f (2x-1)

Function y = log0.1 ^ (6 + x-2x ^ 2) what is the monotone increasing interval
The most important thing is how to find monotone interval~
This is a decreasing function. Why is there a monotonic increasing interval

The function y is a composite function of logo.1u and u = 6 + x-x ^ 2, and logo.1u is a monotone decreasing function
The monotone increasing interval is the monotone decreasing interval of the function u = 6 + x-x ^ 2
For u = 6 + x-x ^ 2
First, 6 + x-x ^ 2 > 0 gets - 2

Given the function f (x) = log a x + 1 / X-1, (A. > 0 and a ≠ 1), the definition domain of the function is obtained, and it is proved that f (x) = log a x + 1 / X-1 is an odd function in the definition domain

The domain D of the function is: {x | X1}
It is proved that: because x belongs to D and - x also belongs to D, f (- x) = log a (- x + 1-X-1) = log a X-1 / x + 1 = log a (x + 1 / x-1) ^ - 1 = - log a x + 1 / X-1 = - f (x), so f (x) = log a x + 1 / X-1 is an odd function in the domain D

If the image of function y = loga (x + 3) - 1 (a > 0, and a is not equal to 1) passes through the fixed point a, if the point a is on the straight line y = (- MX / N) - 1 / N, and Mn > 0, then the minimum value of 1 / M + 2 / N is 0

The image constant crossing point a (- 2, - 1) of the easily obtained function y = loga (x + 3) - 1
-1 = 2m / n-1 / N simplify 2m + n = 1
1/m+2/n=(2m+n)/m+2(2m+n)/n
=2+n/m+4m/n+2
=4+n/m+4m/n
>=4 + 2 √ (n / M) * (4m / N) (take the equal sign if and only if n / M = 4m / N)
=8

The logarithm of the absolute value log of 0 with a as the base (1 + x) is known

Make a difference and discuss the elimination of absolute value
0

Given that f (x) = log, the logarithm of X with a as the base (a > 0 and a is not equal to 1), f (3) - f (2) = 1, find f (x)

f(x)=loga(x)
f(3)-f(2)=loga(3)-loga(2)
=loga(3/2)
=1
a=3/2
f(x)=log(3/2)(x)