Knowing that the function f x kx 2 4x 8 is monotonically decreasing at 4,16, find the range of the v

The axis of symmetry of the function is the straight line x=2 k

If k>0 and f(x) opening is upward, it is a subtractive function to the left of the axis of symmetry. When the axis of symmetry is to the right of the interval [4,16] (which can coincide), the meaning of the question is satisfied. That is, 2 k 16, the solution is 00, and the function is a check function.

The maximum value at (-0) is -2 (a)+2, which is the maximum value obtained when x=- (1 a). To its left is the increasing function and to its right is the subtraction function. The interval satisfying the meaning of the question is - (1 a) -2 to solve a 1 4

If a=0, it is obvious that the function is monotonically decreasing on (- 0), which is not in line with the topic.

If a 0 is an increment function on (- 0) according to the image function, it does not fit the topic.

In summary, a [1 4,+

If a>0, according to the image function, it is a subtraction function on (0,+, which is in line with the title.

If a 0, the function is f(x)=2x, which is an increasing function and does not fit the topic.

If a 0, the function is a checkmark function, and f(x)=2x-a x is at (0,+ on 2 (-2a). is to obtain the minimum value when x= (-a). On its left is the subtraction function and on its right is the increase function, according to the meaning of the question (-a 2) 1, the solution is a -2

In summary, a (-2).

It's possible to miscalculate, and that's the general idea.

If it helps to choose me, thank you.

The derivation yields 2kx-4 =2(kx-2).

When k>0 2 k>16 0k<0 2 k<2 k>1 does not exist.

k=0 is still true so 0<=k<1 8

2.f(x)=ax+1 x+2 is derived from (ax 2-1) x 2 when a>0 1 root a<-2 a<=1 4

a<=0 No, so derivative 2+a x 2, when a<0 -root-a<0 -root>1, so a<=-1, when a>=0 does not exist, so a<=-1

When you're done, choose me.

If k=0, then, is a one-time subtraction function, satisfied. b.If k is not equal to 0, k > 0

then, k*16 2-4*16-81 8; So, 1 81does not exist. Sum up; k=0 or 1 80, a<-2 under the root number, so a>4

3.When a<0, we can know by derivative that -a 2>x <=0, a<=0, so a<0When a>=0, the derivative is unsatisfying aSum up;

k=0f(x)=-4x-8 decrements on r, and k is not equal to 0

then the axis of symmetry x=2 k

If k<0

The opening is downward. then decrements to the right of x=2 k.

So x=2 k is on the left side of the interval.

2/k<=-2

1<=k<0

If k<0

The opening is upward. then decrements to the left of x=2 k.

So x=2 k is on the right side of the interval.

2/k>=10

0 so -1<=k<=1 5

Solution: Image travel method.

Let y=kx 2-4x-8, which is a quadratic function with a quadratic term to be determined, and its image is a discarding and changing line, and its axis of symmetry is x=2 k, if the function f(x) annihilation = kx 2-4x-8 is a monotonic function on [5,20], that is, the quadratic function is a monotonic function on [5,20], so only 2 k 20, or 2 k 5, so 0< k 1 10, or k 2 5 < p >

From the title: when k 0, the symmetry axis fiber is -b 2a=2 k and in the interval [5,20] the single destruction is decreasing by 2 k 20 k 1 10

When k 0 is 2 k 5 i.e. k 2 5 is imitation k 0

When k 0 is given so in sum up k 1 10

Parse. f(5)>f(20)

Because the function is a subtractive function.

Finding the derivative of the function f x kx 4x+8, and the function of f' x 2kx 4 decreasing monotonically, then the rotten brother has f' x 2kx 4 and the silver calendar is positive 0When k=0, it holds;

When k repents 0, x 2 k, 2 k 20, and k 1 10;

When k 0, x 2 k, 2 k 5, the solution is k 0;

In summary, the range of values of the real number k is k 1 10

Hope it helps.

Learning progress o ( o thank you.

If k=0, then y=-4x+8, is eligible. Rulers.

If k>0, then the axis of symmetry is x=2 k, the opening is upward, and in [5,20] monotonically decreasing, it must be: -2 k>=20, getting: k<=-1 10, contradictory.

If k>0, then the axis of symmetry is x=2 and the opening is downward, and the monotonous reduction of [5,20] must be: -2 k<5, obtain: k<-2 5, conform.

Combined: k = 0 or k <-2 5

If the derivative of the function is reversed to f' x kx 4x 8 and the function f' x 2kx 4 decreases monotonically, then there is f' x 2kx 4 0When the dust closure k=0, it is true;

When k 0, x 2 k, 2 k 20, the solution is k 1 10;

When k 0 scatters the brother and repents, x 2 k, 2 k 5, the solution is k 0;

In summary, the range of values of the real number k is k 1 10

1. When k>0, -b 2a=2 k》20 k<1 10 So 02, k<0, -b 2a=2 k"5 k"2 5 contradicts k<0 and is discarded.

3. When k=0, -4x-8=0 decreasing cherry blossom is established.

To sum up, 0"k"1 10

Downstairs, I know that the axis of symmetry is x=2 k, why do you change the song bureau to calculate the band -2 k?

If k>0, then the axis of symmetry is x=2 k, and the opening is upward, and if it is monotonically reduced in [5,20], it must be: 2 k>=20, and get: k<=1 10If k=0, then y=-4x+8, is eligible.

If k<0, then the axis of symmetry is x=2 k, the opening is downward, and in [5,20] monotonically decreasing, it must: 2 k<5, get: k<2 5, conform.

Synthesically, we get the range of values of the real number k (- 1 10].

k=0f(x)=-4x-8 decrements on r, and k is not equal to 0

then the axis of symmetry x=2 k

If k<0

The opening is downward. then decrements to the right of x=2 k.

So x=2 k is on the left side of the interval.

2/k<=-2

1<=k<0

If k<0

The opening is upward. then decrements to the left of x=2 k.

So x=2 k is on the right side of the interval.

2/k>=10

0 so -1<=k<=1 5

1. The function f(x)=kx2-4x-8 is a monotonic function on [5,20], that is, f(x)=kx2-4x-8 is either an increasing function or a subtraction function on [5,20], 2. When k=0, the function is y=-4x-8 is a monotonic subtraction function on [5,20], and when k≠0, the axis of symmetry of the function is x=-(-4) 2k=2 k, as long as the axis of symmetry is x=-(-4) 2k=2 k is not in the interval [5,20] on, i.e. 2 k 5 or 2 k 20

The solution is k < = 1 10 or k > = 2 5, so in summary, k = 0 or k < = 1 10 or k > = 2 5 is actually changed to k < = 1 10 or k >= 2 5.

Solution: f(x)=kx squared - 4x-8

The axis of symmetry of this unary quadratic function image is: x=4 2k=2 k, because f(x)=kxsquared-4x-8 is a monotonic function on [5,20], so [5,20] this interval is on the left or right side of the axis of symmetry, if it is on the left side of the axis of symmetry, 20<2 k, then k>0, 20k<2,k<1 10, get: 02,k>1 10, and if there is no solution, if it is on the right side of the axis of symmetry, 5>2 k

Then when k > 0, 5k>2, k>2 5, get: k>2 5k<0, 5k<2, k<2 5, get: k<0 so the range of the real number k is: k<0 or 02 5

Hope it solves your problem.

(1) When k = 0, f(x) is a one-time function, which is obviously in the monotonic function (2) k != 0, f(x) is a quadratic function, and to make it a monotonic function in [5,20], whether it is monotonically increasing or decreasing, only the image symmetry axis of the function: x = 2 k is not in [5,20].

i.e.: 2 k < = 5 or 2 k > = 20 i.e.: k < 0 or k > = 2 5 or 0 < k < = 1 10;

To sum up: the value range of k is k < = 1 10 or k > = 2 5;

There are two cases:

The first, when k = 0 and f(x) = -4x-8, it is monotonic and qualifies.

Second, k is not equal to 0, f(x) is a quadratic function, only need to consider the position of its symmetry axis, the symmetry axis only needs to be less than or equal to 5 or greater than or equal to 20 to qualify. So 4 (2k) 5 or.

4/（2k)≥20。Solution: 2 5 k 10

In summary: k = 0 or 2 5 k 10

There is no need to consider single increase and single decrease, as long as the position of the axis of symmetry is considered, when k=0 is, this function is linear monotonic and satisfies the situation.

When k<>0, the axis of symmetry x=2 k>=20 or <=5, and k>=2 5 or k<=1 10