It is known that the parabola y = x & sup2; - (K + 1) x + K-2 has two intersections with the X axis. When k = x, the two intersections are symmetric about the origin

It is known that the parabola y = x & sup2; - (K + 1) x + K-2 has two intersections with the X axis. When k = x, the two intersections are symmetric about the origin

The parabola is symmetric about the y-axis, so the axis of symmetry is: x = 0
So according to the formula of symmetry axis, we can know - (- (K + 1)) / 2 = 0, and the solution is k = - 1

When the image of y = sin (Wx + Pai / 4) is shifted to the right by Pai / 4 unit length, it coincides with the image of function y = sin (Wx + Pai / 3), and the minimum value of W is obtained

When y = sin (Wx + π / 4) shifts π / 4 to the right, the expression is as follows:
y=sin(wx+π/4-wπ/4)
According to the meaning of the title, there are
π/4-wπ/4=π/3+2kπ
Obviously, when k = 0, w has a minimum, and the solution is w = - 1 / 12

If the image of the function f (x) = sin ω x (where ω > 0) is shifted to the right by π 4 unit lengths, and the resulting image passes through the point (3 π 4,0), then the minimum value of ω is______ ．

The image of the function y = sin ω x (where ω > 0) is shifted to the right by π 4 unit lengths, and the corresponding function of the image is y = sin ω (x - π 4). Then from the obtained image passing through the point (3 π 4, 0), we can get sin ω (3 π 4 - π 4) = sin π 2 ω = 0, 〈 π 2 ω = k π, K ∈ Z. so the minimum value of ω is 2. So the answer is: 2

If the image of the function y = 1 / 3x + 2 is moved one unit to the right, two units to the down, and then flipped along the straight line y = x, the analytic expression of the image is?

Move the image of the function y = 1 / 3x + 2 one unit to the right and then two units down
So y = 1 / 3x-1 / 3
And then flip along the line y = x to get y = 3x + 1

Take the image of function y = 2 / X along the_____ Axial direction______ Translation______ The graph of function y = x + 2 / 2 can be obtained

Take the image of function y = 2 / X along the____ x_ Axial direction___ Left___ Translation___ 2___ A unit of length,
Let's get the graph of the function y = 2 / (x + 2)

① It is known that sin α cos α = 1 / 8,45 degree

1B
(cosα-sinα)^2=cos^2α-2sinαcosα+sin^2α
=1-2sinαcosα
=3/4
45°

Given that f (x) = 1 / x + LNX, the image of the function g (x) defined on (1,2) is the same as that of F (x)
If a and B belong to (0,2) and satisfy f (a) = f (b), it is proved that a + B is greater than or equal to 2

Let x ∈ (1,2), then 2-x ∈ (0,1)
∵ image of function g (x) defined on (1,2)
It is symmetric to the image of F (x) on (0,1) with respect to the line x = 1
∴g(x)=f(2-x)=1/(2-x)+ln(2-x)
That is g (x) = 1 / (2-x) + ln (2-x)
Let H (x) = g (x) - f (x)
h(x)=1/(2-x)+ln(2-x)-1/x-lnx ,(10
H (x) is an increasing function on (1,2)
Then H (x) > H (1) = 1 + 0-1-0 = 0
That is g (x) > F (x)
Please take the answer and support me

Given the function f (x) = x, G (x) is defined on r even function, when x > 0, G (x) = LNX, then y = f (x) g (x) image is large

According to the monotonicity of function, we can find the extremum of function and monotone interval Y & nbsp; = xlnx & nbsp; (X & gt; 0) & nbsp; & nbsp; & nbsp; & nbsp; Y & # 39; = 1 + LNX, let Y & # 39; = 0 & nbsp; get x = exp {- 1} when 0 & lt; X & lt; exp {- 1} & nbsp; & nbsp; Y & # 39; & lt; 0 & nbsp; & nbsp

Given that the image of the function y = f (x) and the image of y = LNX are symmetric with respect to the line y = x, then f (2)=______ ．

The image of ∵ function y = f (x) and the image of y = LNX are symmetric with respect to the straight line y = x, ∵ f (x) = ex, ∵ f (2) = E2, so the answer is: E2

Given that the image of the function y = f (x) and the image of y = LNX are symmetric with respect to the line y = x, then f (2) =? How to do it?

If the image of the function y = f (x) and the image of y = LNX are symmetric with respect to the line y = x, then f (x) is the inverse function of y = LNX,
Thus f (x) = e ^ x, f (2) = E & # 178;