A couple of math problems are related to functions

Question 1: Won't be embarrassed.

Question 2. In the first case, a=0, then f(x)=2x-3, so f(x)=0, then x= does not belong to the [-1,1] round.

In the second case, a is not equal to 0

The meaning of the title is that f(x) has a solution and at least one solution is on [-1,1].

b^2-4ac》0

f（1）*f（-1）<0

Solution: 1 test If a=1, then f(x)=2x, 2+2x-4, then f(x)=0, then x=-2 or 1 1 is [-1,1] compliant.

If a=5, then f(x)=10x2+2x-8

In this case, f(x)=0 then x=-1 or belongs to [-1,1] conformity.

Solution: Draw the image of f(x) and g(x) in the same coordinate system, and draw f(x) according to the definition, you can see that f(x) has a maximum value and no minimum value.

When x 0, the joint equation is given f(x)=3+2xg(x)=x 2-2x

Get x=2+7 (rounded) or x=2-7

In this case, the maximum value of f(x) is: 7-2 7

Question 2: Divide into unary quadratic equations and unary quadratic equations.

a=0 (unary linear equation).

f(x)=2x-3 is not satisfied.

a！=0 (unary quadratic equation).

The discriminant formula is greater than or equal to 0

f（1）*f（-1）<0

Solve the range. 1

1, I think g(x)=0 or 1, the relationship between f and y and f will be true. From this, the value of x is solved.

2.There are several zero points. f(-1)*f(1)<0 at 1

When 2 times are symmetrical with 1 and -1, the axis of symmetry of this function is x-0, and there is no solution.

Point p(m,n+lnm).

f'(x)=1/x,f'(m)=1/m

So the tangent equation for the image of f(x)=n+lnx at the point p(m,f(m)) is y-(n+lnm)=(1 m)(x-m).

According to the title, 1 m = 1 and n + lnm = 1

The solution yields m=1 and n=1

g(x)=x-1/x-2lnx

g'(x)=1+1 x 2-2 x=(1-1 x) 2x 1, g'(x) 0, g(x) is the increment function.

g(x)min=g(1)=0

g(x) is greater than or equal to 0 constantly.

1:y=-x² +4x

2: When t=, the point p coordinate (, the linear me equation: y=-2x+8

Bring the point p into the line, the equation does not hold, the point p is not on the line me t seconds later, the point p ordinate = t, the point n ordinate = -t +4tpn = -t +3t (0 t 4).

When t=3 2, pn has a maximum value of =9 4

s quadrilateral pncd max = 3 (9 4 + 3) 2 = 63 8

Iterate. xn－x(n-1)＝1－5

x(n-1)－x(n-2)＝1－5

Dan Xin. x3 x2 1 5([2 5]-[1 5])x2 x1 1 5([1 5]-[0 5]) Add up all the above formulas to get you.

xn－x1＝(n－1)－5

n 2009 hours.

x2009－x1＝(2009-1)－5[2008/5]＝2008－5×401＝3

So, x2009 3 1 4

Similarly. yn－y(n-1)＝[n-1)/5]－[n-2)/5]y(n-1)－y(n-2)＝[n-2)/5]－[n-3)/5]y3－y2＝[2/5]－[1/5]

y2－y1＝[1/5]－[0/5]

Add up all the sheds above and get it.

Y1 [(n-1) 5] [0 5]n 2009.

Y2009 Y1 [2008 5] 401 So, Y2009 401 1 402

So, the coordinates of the 2009th tree planting point are (4,402).

Simple, this rule is related to 5.

For the cyclic sequence of xn, which is 1 and 2,3,4,5, xn is required, and only n 5 is needed to take the remainder; The pair of celery collapses in the ordinate yn, and increments by 1 every 5 numbers, that is, the sequence 1,1,1,1,1,2,2,2,2,2,2,..The general term yn=1+[n 5].Bring in n=2009 to get the coordinates.

Because when x=-1, f(x) takes the extremum. then at x=-1, f(x)=0

f (x) = 3ax + 2bx + c, substitute f (-1) 3a-2b + c = 0......（1）

Substituting f(1)=a+b+c=-11......（2）

f（-1）=-a+b-c=5……（3）

Solution (1), (2), and (3) get:

a=1 b=－3 c=－9

So f(x) = x -3x -9x

f′（x）=3x²-6x-9

Where f(x) is taken as an extremum, f(x)=0, i.e., 3x-6x-9=0

Solution: x1 = 1 or x2 = 3

Substituting f(3) = 27-3*9-27 = 27

When x -1, f (x) is greater than 0, so f(x) is incremented at ( 1) (which can be obtained by substituting a number arbitrarily in this interval).

When -1 x 3, f(x) 0, so f(x) decreases at (-1,3).

When x 3, f (x) 0, so f (x) is increasing at (3).

In this way, x=-1 is in an increasing and then decreasing range, so at x=-1, f(x) takes the maximum and f(-1)=5

x=3 is in a decreasing and then increasing range, so at x=3, f(x) takes the minimum and f(3) = 27

The monotonically decreasing interval of f(x) is: (-1,3).

The monotonically increasing interval of f(x) is: (1) (3,).

f(x)=ax3+bx2+cx

f'(x)=3ax2+2bx+c

When x=-1, f(x) obtains an extreme value of 5, which means that f(-1)=5 and f'(-1)=0 i.e.: -a+b-c=5

3a-2b+c=0

a+b+c=-11

>a=1 b=-3 c=-9

f(x)=x3-3x2-9x

f'(x)=3x2-6x-9

Order f'(x)=0, resulting in x=-1 or x=3

f(3)=27-27-27=-27

Therefore, the monotonically increasing interval of f(x) is [negative infinity, -1] and [3, positive infinity], and the monotonically decreasing interval is [-1, 3].

The maximum value is -11 and the minimum value is -27

Convenient Alipay is lonely in my heart.

1。The line y=mx+n intersects y=2x+1 at (2,b) and y=-x+2 at (a,1) to find the value of m,n.

Intersect with y=2x+1 at (2,b), substitution, b=2*2+1=5, b=5, intersect with y=-x+2 at (a,1) to bring in, 1=-a+2,a=1, that is, the straight line y=mx+n passes through (2,5) ,1,1)5=m*2+n,1=m*1+n

5-1=2m+n-1m-n,4=m1=4+n,n=-3,2.When k is what the value, the function y=2-x, y=-x 3+4, y=4 k x-3

1 synonymously obtains two systems of equations, brings in the intersection point, and obtains two equations about m and n, i.e., 2m+n=5 and m+n=1 to solve m=4, n=-3

I didn't understand what question 2 meant.

1, solution, substituting (2,b) into the function y=2x+1 to get b=5, the same way gets a=1, so the straight line y=mx+n passes through the points (2,5) and (1,1) that is, 5=2m+n, 1=m+n, and gets m=4,n=-3 What does the second question ask? 2 solution, the simultaneous equation y=2-x, y=-x 3+4 gives the intersection point of the three function images as (-3,5), and brings this point into the equation y=4 k x-3 to get k=-3 2

Because y=mx+n, y=2x+1 intersect, the two linear equations are combined: mx+n=2x+1, that is, x=(2-m) (n-1)=2; In the same way, the equation 2-y=(y-n) m, y=1, can be calculated! Let's do the math yourself

The second question seems to be incomplete......

1 Substituting (2,b) into y=2x+1 2 and finding the solution of y=2-x,y=-x 3+4 equations.

b=5 x=-3,y=5

Substitute (a,1) with y=x+2 and (-3,5) with y=4 k x-3

y=4/k x-3a=1 k=-3/2

Substitute (2,5),(1,1) into y=mx+n

5=2m+n

2=m+nm=3n=-1

(1) Because l1:y=2 3x+8 3 intersects with the line l2:y at the point c, so 2 3x+8 3=-2x+16 x=5 y=6, so c(5,6).

So the height of the triangle is 6

When y=0.

l1 2/3x+8/3=0 x= -4l2 -2x+16=0 x=8

So a(-4,0)b(8,0)ab=12

Triangle area = 1 2 times 6 times 12 = 36

2) Because b(8,0), the abscissa of d is 8

So y=2 3 times 8+8 3=33 12 so d(8,33 12) so db=33 12

So the ordinate of e is 33 12 so 33 12 = -2x+16 x= so e(,33 12).

So de= db=33 12 de=

Ideas: (1) First, the coordinates of points A, B and C are required, and then the area of the triangle ABC is calculated according to the coordinates of the three points.

2) According to the coincidence of point g and point b and y=2 3x+8 3, determine the coordinates of point d, and then determine the coordinates of point e according to the coordinates of point d and y=-2x+16, and then determine the length of de and ef.

Haha, I haven't studied mathematics for two years, and this problem is still very easy to solve (1)2 3x+8 3=-2x+16 to get the coordinates of point c (5,6), and then calculate a(-4,0), b(8,0).

Area = 1 2c to the distance from the x-axis xab = 1 2x6x12 = 36 (2) d and e are in the same straight line, so d(8, x) is substituted into the formula to get d(8,8), and then e(4,8),f(4,0).

So to de=4, ef=8

It's not easy to type formulas, so I'll briefly describe it, but the solution is absolutely correct, don't worry, just convert him into mathematical form.