High School Simple Function Problem, A Simple High School Function Problem

Question 1: Because f(x)=x 5+ax 3+bx-8, f(-2)=-2 5-a·2 3-2b-8, because f(-2)=10

So -2 5-a·2 3-2b-8=10, so 2 5+a·2 3+2b=-18

And because f(2)=2 5+a·2 3+2b-8, f(2)=-18-8=-26

Problem 2: Because f(x)=x 2+2(a-1)x+2, the derivative of the function is g(x), so g(x)=2x-2(a-1) =2x+2a-2 because the function f(x)=x 2+2(a-1)x+2 is a subtraction function over the interval (- 0), so when x(-0), g(x) 0

i.e. 2x +2a - 2 0

So x 1-a

Because x (-0).

So 1-a 0

Solution A 1

1. Prompt: Replace it with the whole.

2. As long as the axis of symmetry (1-a) 0 is sufficient.

g（x）=x^5+ax^3+bx

f（x）=g（x）-8

So g(-2)=f(-2)+8=18

g(x) is an odd function, so g(2)=-18

f(2)=g（2）-8=-26

The axis of symmetry of the image is to the right of the y-axis.

1-a>0,a<1

The axis of symmetry of the function f(x) is x=a, and considering several divisions of a, it is actually considering the position of the axis of symmetry The axis of symmetry is on the left side of the interval, then f(x) is monotonically increasing on [-2,2] The axis of symmetry is on the right side of the interval, then f(x) is monotonically decreasing on [-2,2] The axis of symmetry is between the interval [-2,2], then only the minimum value of the function can be determined, that is, f(x) has a minimum value when x=a, but its maximum value cannot be determined, because the left side of the symmetry axis decreases, Incrementing on the right, you can't determine the size of f(-2) and f(2), i.e. you can't determine the maximum value of the function.

Therefore, it is necessary to subdivide it further, the middle point of -2 and 2 is 0, then it is divided into (-2, 0) and [0, 2) two ends, so that the symmetry axis is in a certain interval according to the symmetry to know the size of f(-2) and f(2), for example, when the symmetry axis is in (-2, 0), then x=2 is far away from the symmetry axis, at this time it is f(2) f(-2).

One: If the axis of symmetry x=a does not fall on [

interval, the function is monotonic, but when the axis of symmetry is on the left of a given interval, it is monotonically increasing, the minimum is x=-2, and the maximum is taken at x=2; When the axis of symmetry is on the right side of a given interval, the function is monotonically decreasing, with the maximum at x=-2 and the minimum at x=2.

2: If the axis of symmetry x=a falls on [

interval, then the minimum value of the function must be taken when x=a, but where to take the maximum, it will be divided into two types of cases When a is from -2 to 0, then 2 is farther away from a than -2 from a, and the maximum value of the function is taken at x=2; When a is from 0 to 2, -2 is farther away from a than 2 from a, and the maximum value of the function is taken at x=-2.

To sum up, A should be considered in four categories.

Swap x with 1x.

2f(1 x)+f(x)=3 x

There is another 2f(x)+f(1 x)=3x

Solve a system of binary linear equations.

f(x)=2x-1 x

Define the domain as.

The range is or all the fractions that have occurred are his range.

So the range is.

Define the range as (0,5) and the range as (0, out of a score) What is the maximum score of the range, and the upper limit of the range is that number.

1. When x 2, f(x)=|x-1|+|x-2|=x-1+x-2=2x-3≥1≥a

When 1 x 2, f(x)=|x-1|+|x-2|=x-1+2-x=1≥a

When x 1, f(x)=|x-1|+|x-2|=1-x+2-x=3-2x≥1≥a

Therefore, a 12, the function y=3x to the left and the ant to translate 1 unit, that is, the axis of symmetry to the left to translate one unit, to get the number clear.

y=3(x+1), and then translate down by 1 unit, that is, the vertex translates down by 1 unit, obtain.

y=3x（x+1）²-1

3. Regard x+4 in the function y=f(x+4) as a whole, and the image of the function f(x) passes through the point (0,1), that is, when x+4=0, y=1, that is, the image of the function y=f(x+4) passes through the point (-4,1).

The first rational rubbers are the sum of the distances from the points on the x-axis to 1 and 2. Obviously, when located between two points, the distance and the minimum starvation are 1

The second one is 3 (x+2) — the length of the downward translation.

The third overall reincarnation early element treats x+4 as x to get (-4,1).

a=+/1

When a=-1: f'(x)=3ax^2+2bx+c=-3x^2+2bx+c

Order f'(x)=0.

3x^2+2bx+c=0

Since the function is decreasing on (1,2), so f'(x) is less than 0 on (1,2).

When a is less than 0, the derivative cannot make the single compartment interval (1,2). Draw the picture below. A graph of the derivative function.