Please help me analyze this math problem high school .

f(2x+3) defines the domain as x.

The domain of f(2x+3) is (-4,5), i.e., -4 is the domain of f(x) is: (-5,13).

So there is: -5<2x-3<13, and the solution: -1The domain of

The definition field is all about the range of x values, which means that the condition is equivalent to telling you x (-4,5), then 2x+3 (-5,13).

The characteristic of the function is that for f(z), no matter what form z is, the conditions that need to be met are the same; That is, 2x-3 must be (-5,13) to solve x (-1,8).

f(2x+3) defines the domain as x.

4-5<2x-3<13, corresponding to -1f(2x-3) definition domain (-1,8).

For example, f(2x+3)=2x+3, and the domain of f(2x+3) is the domain of the function 2x+3, which is x.

Let's synthesize it yourself, it's not so troublesome, the definition domain refers to the variable, and the function code f has a meaning condition, which is the specified range of parentheses.

It's all about x, not 2x+3, you know.

Solution: (1) 丌 2 "乄 "丌, p point on the final edge of the corner (x,4), then 丨op丨= (x 2+4 2)

cos乄=x 丨op丨=1x 5

x=-3, (x=3 rounded).

tan 乄 = -4 3

2) The final point of the q angle p(x,-1).

tanq=-1 x=-x,x= 1,丨op丨= [x 2+(-1) 2]= (x 2+1)= 2

When x=1, sinq=-22, cosq=22;

When x=-1, sinq=- 2 2, cos=- 2 2

According to the definition, it is OK to make a right triangle through this point, and then according to the range of x, it is a negative number, so as to solve this problem, it is not difficult.

(1) Since cos = x 5 and because p point is (x, 4), then sin = 4 5

And because in the second quadrant, we know x<0 and thus x=-3 so tan =4 -3=-4 3

2) Because the angle crosses the point (x, -1), then tan = -1 x and tan = -x

Then we get -1 x=-x, we get x=1, then we get when x=1, sin =- 2 2, cos = 2 2, when x=-1, sin =- 2 2, cos =- 2 2,

Are you sure about high school questions? It's a bad one.

Because the distance between the two intersections is 3, let the two solutions be x1, x2 and x1 > x2 respectively

x1-x2=3

x1+x2=-m

x1x2=m+3

x1+x2)^2-4(x1+x2)=(x1-x2)^2m^2-4(m+3)=9

Just solve the quadratic equation, and I won't be noisy.

Proof: Analysis, this problem only needs to prove min[f(-2+1 2 (n-1)]>max,n n+,x (0,e].

When a=-1, f(x)=-x-ln(-x),x [-e,0).

Derivative f'(x)=-1-1/x=-(x+1)/x

then ln(-x)f'(x) (x+1)]+1 2=-[ln(-x)] x+1 2=g(x)(credit)(x<0).

Derivative g'(x)=[ln(-x)-1] x, let g'(x)=0 gives the unique stationing point x=-e, when x (-e) has g'(x) > fierce coarse 0, knowing that g(x) increases monotonically on (- e);

When x (-e,0) has g'(x)<0, knowing that g(x) decreases monotonically on (-e,0), and obtains a maximum value at x=-e, the maximum value must be its maximum.

then x [-e,0), there is g(x) maxg(x)=g(-e)=1 e+1 2 "Song Zhi Tremor 1 2 + 1 2=1....1)

Next, the function h(x)=ln(x+1)-x,x 0 is introduced, and the derivative h is obtained'(x)=-x (1+x)<0, x>0, knowing that h(x) decreases monotonically on x>0.

then h(x) loses 0 in the wild, and gets ln(x+1)0....

Let's take 1-1 2 (n-1)(>0) instead of the above equation x has ln[2-1 2 (n-1)]=ln<1-1 2 (n-1).

-ln[2-1 2 (n-1)]>1-1 2 (n-1)].

Then f[(-2+1 2 (n-1)]=2-1 2 (n-1)-ln[2-1 2 (n-1)]>2-1 2 (n-1)-[1-1 2 (n-1)]=1....2)

Synthesis (1), (2) yields f[(-2+1 2 (n-1)]>1>g(x)=ln(-x)f'(x)/(x+1)]+1/2

Thus for any n n+,x[-e,0)there is always f[(-2+1 2 (n-1)]>ln(-x)f'(x) (x+1)]+1 2 is true, and the proposition is proven.

Set the midpoint (x,y).

then there is (2x-4, 2y+2) on the circle.

2x-4)^2+(2y+2)^2=4

x-2) 2+(y+1) 2=1 is the midpoint trajectory.

b^2+c^2-a^2=2bccosa

c^2-3√5c+10=0

The solution yields c=2, 5, or 5

It can be seen that the trilateral relation of the triangle is satisfied

A vs. C?

There are two answers?

It is clear that tana = 3 3

And a b = 5 15 = 3 3, so c 90° so c = 2 5

The known condition is the edges and corners, and such conditions are not unique, so there are two possibilities for the length of c.

sinb = sina * b a = 1 2 * 15 5 = 3 2 b = 60 degrees or 120 degrees.

If b=60 degrees, then c=90, c=a sina *sinc=2 5 If b=120 degrees, then c=30 degrees, c=a sina *sinc= 5 choose c

1) From the meaning of the title, we can know that f'(x)=cosx-xsinx, so that f'(x)=0, we can know cosx=xsinx, and we can know x (0, 2).

2) Then the extreme values in the question are arranged from small to large, so an+1-an>0, exclude a

3) With special values, when x 6 and x 3 there will be two extreme points, then the difference between the two extreme points (0, 2).

Therefore, choose B