Solve High School Math Problems!! I want the process.

f(x) is obtained by finding the derivative'=3x^2-2x

Req f(x).'=0, x=0 or 2 3,f(x).'>0, x>2 3 or x<0

f(x)'<0, 0 f(-3)=-36

Therefore, the range of the function on [-3,2] is [-36,4], which is calculated quickly, I don't know if there is anything wrong.

The general idea is this, by verifying the monotonicity of the function (subregional) to compare the value of the function, for example, in the negative infinity to zero increment, 0 to 2 3 decreasing, you have to compare the lowest point of the value (minimum) f(2 3) and f(-3) size, then there is f(-3) is the smallest in the defined domain; After that, the function increases, and after 1, Evergrande is 0 and is an increasing function, and the intrinsic f(2) obtains the maximum.

Derivative, f'(x)=3x²-2x

Solution f'(x)=3x -2x=0, get, x=0 or x=3 2 solution f'(x)=3x -2x<0, 0 calculates the value of 4 points, and these 4 points determine the value range.

f(-3)=-36 f(0)=0 f(3 2)=9 8 f(2)=4, so the range is [-36,4].

Solution: Finding its range on [-3,2] is actually finding the maximum value on this interval.

Because: f'(x)=3x 2-2x=0 x=0 x=2 3 Find the possible extreme points.

The maximum value must be obtained at the extreme point or endpoint.

f(-3)=-27-9=-36

f(0)=0

f(2/3）=8/27-4/9=-4/27f(2)=8-4=4

So f(x)=x -x has a minimum value of f(-3)=-27 on [-3,2] and a maximum value of f(2)=4

f(x)=x -x is found in the range of [-3,2] as: [-27,4].

The monotonic interval of this function can be found by the derivative method, increasing on [-3,0], decreasing on [0,2 3], and increasing on [2 3,2], so the range is [-36,4].

Deriving this equation gives f'(x)=3x -2x, and then judges its monotonicity, and thus determines the trend of increasing and decreasing, and we get the maximum value, and the middle of the two plants is the value range.

Known; f(x)=x+a/x^2+bx+1

1<=x<=1) is an odd function! Seeking a, b? (2) Judge the monotonicity of f(x) and prove it with definitions! Ask for the process, be detailed!

1) Because of the odd function, the origin must be crossed.

f(0)=0

So we get a=0

f(x)=x/(x^2+bx+1)

Because the function is odd.

So f(-x)=-f(x); This results in -x (x 2-bx+1) = -x (x 2+bx+1).

Therefore, b=02)f(x)=x (x 2+1) deformation gives f(x)=1 (x+1 x) so that g(x)=x+1 x

g(x) in.

0, 1) on monotonically decreasing.

and g(x) in.

0,1) is greater than 0

f(x)=/1g(x)

So f(x) increases monotonically over (0,1).

When the first question is not continuous x 0, f(x)=1, when x 0, f(x)=-1, the second question is the inverse proportional function y=1 x, and the image can be obtained by translating the abscissa to the right by 2 units, and the discontinuity can be seen on the image.

In the third question, y=x -6x+9 (x-3)=x-3 (x≠3)x=3, the y=2 break is x=3

Solution: (1).

y=sinx, according to the trigonometric properties, can be written as:

sin(x+ )=-sinx=sin(-x) Therefore, y=sinx has a "p(a)" property.

According to the periodicity of the trigonometric function, we get:

sin[x+(2k+1) ]=sin(-x), where k is an integer, a=(2k+1), where k is an integer.

2) According to what is known:

f(x)=f(-x)

When x 0, f(x) = (x+m).

Let x 0, then:

f(x)=f(-x)=(-x+m)²

Therefore: (x+m) = (-x+m).

i.e.: |x+m| = |-x+m|

x+m=-(-x+m)

x+m=-x+m

Get: 2x=0, which is not in line with the topic, because x also makes sense if it is less than zero m=0 so:

f(x)=x²

When x [0,1], obviously the maximum value y=f(x)=1 is taken when x=1

(1) Determine whether the function y=sinx has "p(a) property", and if it has "p(a) property", find the value of all a; If it is not of a "p(a) nature", please explain the reasons.

f(x+a)=f(-x)

sin(x+a)=sin(-x)=-sinxa=(2k+1), k is an integer.

2) It is known that y=f(x) has "p(0) property", and when x is less than or equal to zero, f(x)=(x+m) squared, find the maximum value of y=f(x) on [0,1].

y=f(x) has a "p(0) property".

then f(x) = f(-x).

f(x)=(x+m) x 0

Then f(x)=(x+m) x 0 when x is on [0,1].

1.If -m 0

f(x)max=f(1)=(1+m)²

2.If 0<-m<1

f(x)max=max=max;

1.If -m>10

f(x)max=f(0)=m² 。

(1) sin(x+a)=sin(-x)=sin(x+ +2k) because x is arbitrary, so answer a=(2k+1).

2) p(0) property: f(x)=f(-x), i.e. even function property f(x)=(x+m) 2 x<=0 is the axis of symmetry x=-m, the opening is upward, and the parabola with vertices on the x-axis is on the left part of the y-axis.

It is necessary to divide the axis of symmetry -m<=-1 2 m>-1 2, and draw a diagram to help understand -m<=-1 2, max=f(0)=m 2-m>-1 2, max=f(1)=f(-1)=(-1+m) 2

Since sin(-x)=-sinx, if there is such a property, then sin(x+a)=-sinx, so a=k. k is an odd number

The second question has a property that obviously leads to the fact that f is an even function, so that the analytic formula of the function comes out, and then it can be solved.

Answer: Let the common ratio of the proportional series be q

a1*a3=a2²

a1a2a3=8

a2³=8a2=2 a1+a2+a3=7

1) a1+a3=5

2) a2/q+a2*q=5

2/q+2q=5

2q²-5q+2=0

q-2)(2q-1)=0

q = 2 or q = 1 2

q>1

q=2 a1=1

an=2^(n-1)

s8=a1(1-q^8)/(1-q)=2^8-1=255

As can be seen from the question: a2=q*a1; a3=q^2*a1;

a1+a2+a3=a1(1+q+q 2)=7 (1)a1a2a3=(a1*q) 3=8 So q*a1=2 is substituted for (1).

It can be obtained: (1+q+q 2)*2 q=7 i.e.

2q^2-5q+2=0

q=2 or q=

Because q 1 so q = 2 a1 = 1

The rest is fine

Hope it helps

Use the commutation method. Think of sinx+cosx as a new dollar. If there are no special requirements, the domain of x is considered to be r.

Then the range of the new element is the positive and negative root two closed intervals. (Needless to say, the oblique formula) sinx*cosx is replaced by (sinx+cosx) -1 2, and then it is the problem of the evaluation domain of the unary quadratic equation.

1=2a+3b>=2*under the root number (2a*3b), so 2a*3b<=1 4

ab<=1 24, so the maximum value is 1 24

The maximum value is taken if and only if 2a=3b, i.e., a=1 4, b=1 6.

Downstairs is the right solution;

Bias: a=(1-3b) 2

a*b= -2/3b*b+1/2b

The function -2 3b*b+1 2b, with the opening facing down, is the maximum value of the solution bmax=(b1+b2) 2=1 6

i.e. a*b(max)=-1 24+1 12=1 24

Using b as the y-axis and a as the x-axis, the function image is obtained as above.

It can be seen that (a, b) are points on a straight line, and the product of ab is the area of the following figure.

To make ab the maximum, choose the midpoint of the straight line, i.e. a: 1/4, b: 1/6Just multiply.

Analysis: If there is a real root, m 2-4*2*n>=0....1) 2, m, n are the first three terms of the difference series, then 2m=2+n....

2) From (1) and (2), m<8-4 sqrt(3) or m>8+4sqrt(3) sqrt means the root number.

Tolerance d=m-2

So d<6-4 sqrt(3) or d>6+4 sqrt(3).

Solution: =m 2-4*2*n>0, 2m=2+n, that is, n=2m-2, substituting the first formula, get m>8+4 root number 3, m<8-4 root number 3, so d = m-2, so d > 6 + 4 root number 3, d < 6-4 root number 3

x>=0……Type 1.

x+y-3<=0……Type 2.

x-2y<=0……Type 3.

Multiply 2 by 2 and add 3 to get:

x<=2……4 formulas, combined with 1 formula, get.

0<=x<=2……Type 5.

It is available from 3 formulas.

y>=x/2

It can be obtained by 2 formulas.

y<=3-x

That is, 0<=y<=3-x......Type 6.

Then there is x+2y>=0+2*0=0

x+2y<=x+2*(3-x)=6-x……Equation 7 is obtained from equation 5: -2<=-x<=0......Type 8.

Formula 7 plus formula 8 can be obtained:

x+2y<=x+2*(3-x)=6-x<=6, so 0<=x+2y<=6