What is the maximum value of x k x 2?

Find the maximum value of the function f(x)=x (k+x).

Solution: Define the domain: x≠-k

Order f'(x)=[(k+x) -2x(k+x)] (k+x) =(-x +k) (k+x) =0, get x =k

Therefore, the station point x= (k)=k [let k>0] [x=-k be rounded]; When x0; When x>k f'(x)<0；

So x=k is the maximum, and the maximum of f(x) = f(k) = k (4k) = 1 (4k).

x➔-klim[x/(k+x)²]=-∞

x➔+∞lim[x/(k+x)²]=x➔+∞lim[x/(k²+2kx+x²)]=x➔+∞lim[1/(k²/x+2k+x]=0

x➔-∞lim[x/(k+x)²]=x➔-∞lim[x/(k²+2kx+x²)]=x➔-∞lim[1/(k²/x+2k+x]=0

Therefore, when k > 0, the maximum value of the function is its maximum value of 1 (4k).

Solution: Order: x (k+x) 2-l

Simplified: lx 2+(2kl-1)x+k 2l=0=1-4kl 0

kl≤1/4

Then, according to the positive and negative properties of k, the range of l is obtained, that is, the maximum value of x (k+x) 2.

At k>0, there is a maximum value of 1 4k

At k<0, there is a minimum of 1 4k

k=0, x (k+x) 2=1 x is infinitely close to 0

There are several situations:

1) When k=0, there is no extremum.

2) When k is not equal to 0, x (k+x) 2=(k+x-k) (k+x) 2= -k (k+x) 2+1 (k+x)=-k[1 (k+x)-1 2k] 2+1 4k

At k>0, there is a maximum value of 1 4k

At k<0, there is a minimum of 1 4k

When k = 0, there is obviously no extremum.

When k is not equal to 0, x (k+x) 2=(k+x-k) (k+x) 2= -k (k+x) 2+1 (k+x)=-k[1 (k+x)-1 2k] 2+1 4k

At k>0, there is a maximum value of 1 4k

At k<0, there is a minimum of 1 4k

k>0, the opening is upward.

f(x)=k(x+1 k) 2+1-1 kk>0, the opening is upward.

The axis of symmetry x=-1 k,k>0, so -1 k<0 if -3<=-1 k<0,0<1 k1 3

Then x=-1 k, the first beat is small=1-1 k=-4, 1 k=5, k=1 pei hall 5, does not meet k>1 3, and is not true.

If -1 kk>3, so 0 so x=-3, minimum = 9k-6+1=-4, k = 1 9<1 3, the symbol is matched with celery.

So k=1 9

Summary. Hello, glad to answer for you. x -2x-k>0, x (1,2), find the range of k (-2) 4 1 (-k) 04 4k 04k -4k -1

x -2x-k>0, x (1,2), find the range of k.

Good. Hello, very orange book is happy to answer for you. x -2x-k>0, letter touch x (1,2), find the range of k (-2) Slide talk 4 1 (-k) 04 4k 04k -4k -1

Are you sure? x may not be 2.

Isn't x possible to be any number from 1 to 2?

From the known: x -2x-k>0, x (1,2), this problem examines the discriminant formula of the former good root, and the key to solving the problem is to obtain the unary one-dimensional inequality about k This problem belongs to the basic problem, which is not very difficult, and when solving this type of problem, according to the number of roots combined with the discriminant formula of the root, the Huiyan lead equation (system of equations, or inequalities) is the key to the source of jujube.

It's the same, just a substitute number.

Oooh oh. Because it's incremental, isn't it? So this inequality must be solved by the largest number in the set.

To find the solution set of inequalities, you can first represent the solution set of each inequality on the number line and observe the common part. Then remove the brackets, move the terms, merge the similar terms, and turn the coefficients into a moment to notice whether the bottom pit balance is divided by a positive or negative number.

2x square + kx + 7

2 (x square + k 2 + 7 2).

2 ( square + k 2x + (k 4) square - (k 4) square + 7 2) = 2 ((x + (k 4) square + 7 2 - (k 4) square) ) = 2 (x + (k 4) square + 7-2 (k 4) square, when x = - (k 4) square, the original formula has a minimum value of 2, that is.

7-2 (k4) square = 2

2 (k 4) square = 5

k4) squared = 5 2

k/4=±√(5/2)

k=±4√(5/2)

Write the Heng Ke function as y=k(x+1).

1-k, so the axis of symmetry is x=-1

1) If k 0, the parabola opening is towards the finch and the maximum value of 3 is taken when x=2So k=1 4

2) If k 0 and the parabolic opening is downward, take the maximum value of 3 when the axis of symmetry is x=-1So k=-2

So k is 1 4 or -2

To solve this kind of problem, the idea of categorical discussion is mainly considered; To find the maximum value of a function over a given interval is to discuss monotonicity.

to the original function. The nature of the property should also be grasped thoroughly).

The function y=-x +m

x+2 axis of symmetry.

The relative position of x=m 2 and the interval [0,2] is the key to solving the problem.

1) If the axis of symmetry is on the left side of the interval, then f(x)max=f(0)=2;

2) If the axis of symmetry is on the interval, then f(x)max=f(m)=2;

3) If the axis of symmetry is located on the right side of the interval, then f(x)max=f(2)=2m-2;

At this point, k = 2, or k = 2m-2, so the next guess is to discuss the relationship between the size of 2 and 2m-2: (1) If m = 2, then k = 2;(2) If m>2, then k=2m-2; (3) If m<2, then k=2

Keep talking!

y=-x2+mx+2, and the axis of symmetry is x=m2

Knowing 0<=x<=2, the value of m is discussed in segments.

When 0=4, the symmetrical axis x>=2,x

4 to get the maximum.

k4m-14;

When m is greater than the family car return 4, the function is an increasing function in the interval [0, sail missing 2]. When x=2, y=m-2. So when the value of m is positive infinity, the maximum value on 0 megastarvation x 2 is m-2 (when x = 2) and the value of k is positive infinity.

This problem depends mainly on the grasp of the axis of symmetry and monotonicity of the quadratic function.

Axis of symmetry: x=-b 2a=-1

Vertices: (1, 4).

The image opens upwards, decreasing to the left and increasing to the right.

So, when k+2<-1, i.e., k<-3.

The maximum value is obtained at k+2, and the function of k+2 generation is obtained as: k 2+6k+5 when k >-1, the maximum value point is obtained at k, and k is substituted in, the maximum value is: k 2+2k-3 when 3<=k<=-1, the interval is distributed around the axis of symmetry, and the maximum value point is the vertex, that is, -4

f(x)=x +2x-3=(x+1) -4 when k -3, f(x) [f(k+2),f(k)];

When -3 k -2, f(x) [4,f(k)];

When -2 k -1, f(x) [4,f(k+2)];

When k -1, f(x) [f(k),f(k+2)];

To classify the discussion, the position relationship between the axis of symmetry x=-1 and the interval [k,k+2] is to be discussed. Three cases: 1. The axis of symmetry x=-1 is on the left side of the interval [k,k+2]; 2. The axis of symmetry x=-1 is in the middle of the interval [k,k+2]; 3. The axis of symmetry x=-1 is on the right side of the interval [k,k+2].

Try it yourself.