A few questions about functions, a few questions about functions

1. Isn't the even function about y-axis symmetry? Why is it also about symmetry of the coordinate origin??? It's just that I don't understand this!!

So-called"About coordinate origin symmetry", does not refer to the function, but refers to the definition domain ---that is, the range of values of x, for example, -11) In the definition of odd functions and even functions, it is required that the interval corresponding to the definition domain of the function is symmetrical with respect to the coordinate origin, and if the interval corresponding to the definition domain of a function is asymmetrical with respect to the coordinate origin, this loses the necessary condition that the function is an odd function or even function, and the function has no parity at all.

That's right. odd function f(-x)=-f(x); The even function f(-x) = f(x), so defining the corresponding interval of the domain is symmetrical with respect to the coordinate origin, which is the basic requirement of the odd function or even function.

2) The definition domain of a function is a necessary condition for the function to be even, but not sufficient.

That's right. Defining domain symmetry with respect to the coordinate origin is not necessarily an even function.

3) A function being an even function is a necessary but not sufficient condition for the symmetry of the coordinate origin of its definition domain.

Mistake. The assertion could be changed to (2).

4) The sufficient and necessary condition for the image of a function to be symmetrical with respect to the coordinate origin is that the function is an even function.

Mistake. The image of a function is symmetrical with respect to the coordinate origin, then f(-x)=-f(x)The argument is changed to say that the sufficient and necessary condition for the symmetry of the image of a function with respect to the coordinate origin is that the function is an odd function.

5) The sufficient and necessary condition for the image of a function to be symmetrical on the y-axis is that the function is even.

That's right. The image of a function is symmetrical with respect to the y-axis, then f(-x)=f(x)

2. f(1)=0

f(f(f(f(－1)=-2

f( on interval (5,5).

When -5f(x)=-5

When -4f(x)=-4

When -3f(x)=-3

When -2f(x)=-2

When -1f(x)=-1

When 0f(x)=0

When -1f(x)=-1

When -2f(x)=-2

When -3f(x)=-3

When -4f(x)=-4

So: the image of the function is step-like.

Solution:1∵ f(x)+lgxf(1/x)=x+1①

f(1 x) lg(1 x)f(x) 1 x 1, simplified to f(1 x) lgxf(x) 1 x 1

Synact is seen as a system of binary linear equations with respect to f(x), f(1 x), and f(1 x) is subtracted

The solution yields f(x)=(x1)(x-lgx) x(1 lg x).

f(10)＝(10+1)×（10-lg10）／10×（1＋lg²10）=99/20

2.∵f(x)+2f(-x)=x²＋x＋1

f(-x)＋2f(x)＝x²-x＋1

Synovia is regarded as a system of equations about f(x) f(-x) and subtracts f(-x).

The solution yields f(x) 1 3 x 1 3

3.∵f(x)+2f(1-x)=x²＋x＋1

Replace 1-x with x to get f(1-x)+2f[1-(1-x)] 1-x) +1-x+1

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

The two formulas are combined, and f(1-x) is eliminated.

f(x)=1/3x²－7/3x-7/34.

Solution: (1) Bring x=10 in: f(10)+f(1 10)=11 Bring x=1 10 in: f(1 10)-f(10)=11 10

Solve the equation to obtain: f(10)=99 20

In the same way, bringing x=1 x into the equation yields the expression f(x) that can be solved by combining x(1 x)+lg(1 x)f(x)=1 x+1 with f(x)+lgxf(1 x)=x+1.

Explanation: With the inspiration of this question, I think you should have good ideas for the following questions, I hope you can have good ideas, hehe.

These questions are not very difficult, and it is recommended that you look at the calendar questions that are very simple.

Let f(x)=ax+b be substituted to get 3(ax+a+b)-2(ax-a+b)=2x+17, and split it to get (3a-2a)x+3a+4a+3b-2b=2x+17, that is, ax+7a+b=2x+17a=2,b=3, so f(x)=2x+3

f(2)=0 means that Fu Chunmeng is the independent variable x=2 of the bridge function f(x), and the function ax 2+bx=0 Therefore, bringing x=2 into the equation is 4a+2b=0

f(x)=(a^x-1)/(a^x+1)

f(-x)=[a^(-x)-1]/[a^(-x)+1]=(1-a^x)/(1+a^x)

(a^x-1)/(a^x+1)

The f(x) function is an odd function.

f(x)=(a^x-1)/(a^x+1)

1-2/(a^x+1)

a>1 a x monotonically ascends, a x+1 increases monotonically, and 2 (a x+1) decreases monotonically.

f(x) monotonically increasing.

f(x)=(a^x+1-2)/(a^x+1)=(a^x+1)/(a^x+1)-2/(a^x+1)

1-2/(a^x+1)

a x>0, so a x+1>1

So 0<1 (a x+1)<1

2<-2/(a^x+1)<0

1-2<1-2/(a^x+1)<1+0

So the value range (-1,1).

f(-x)=(a^-x-1)/(a^-x+1)

Multiply a x up and down, and a x*a -x=1

So f(-x)=(1-a x) (1+a x)=-(a x-1) (a x+1)=-f(x).

Define the domain because the denominator a x+1>1 is not equal to 0

So the definition domain is r, symmetry with respect to the origin.

So it's an odd function.

a>1, then a x is an increasing function, a x+1 is an increasing function, 1 (a x+1) is a decreasing function, -2 (a x+1) is an increasing function, and f(x)=1-2 (a x+1) is an increasing function.

Monotonicity can also be demonstrated in this way:

f(x)=(a x-1) (a x+1)=1-2 (a x+1), x belongs to r, take x1, x2 belongs to r, and x11, so a x1f(x1)-f(x2)=2 (a x2+1)-2 (a x1+1)=2[(a x1+1)-(a x2+1)] [(a x1+1)(a x2+1)]=2(a x1-a x2) [(a x1+1)(a x2+1)]<0, i.e. f(x1) < f(x2), so f(x) is an additive function over r.

The odd function is reduced to f(-x)=-f(x).

Increase function Let x1>x2 use subtraction to pass the denominator horizontally"0 The numerator is 2 times ax1 times subtract ax2 because a is everstable in 1, so the numerator is greater than 0, so f(x1)-f(x2) is greater than 0 is the increase function.

Since f(x) is a one-time function, let f(x) ax+b be 3f(x+1)-2f(x-1)=2x+17 to get 3*(a*(x+1)+b)-2*(a*(x-1)+b) 2x+17

Simplify, we get ax+5a+b 2x+17, because f(x) is a function, then a 2, and then bring a 2 in, we get b 7

Therefore f(x) 2x+7.

Pay attention to do more questions and dare to make bold assumptions.

Directly let f(x)=kx+b, then the inscription: 3[k(x+1)+b]-2[k(x-1)+b]=2x+17

Then make the left one to the simplest: kx+5k+b=2x+17 and then it corresponds one-to-one: k=2, 5k+b=17

So b=7, so the primary function is: y=2x+7

Order a>b

So f(a) = f(b+a-b) = f(b) + f(a-b) due to a-b>0

So f(a-b)<0

So f(a) so f(x) monotonically subtract function.

f（0+0）=f（0）+f（0）

So f(0)=0

Again, by f(x-x) = f(x) + f(-x).

i.e. f(x) + f(-x) = 0

So f(x) is an odd function.

So |f（x）|is an even function.

is monotonically subtracted by f(x), f(0)=0

Draw your own picture|f（x）|

It's easy to know.

There is no solution when a is less than 0.

A solution when a is equal to 0.

2 solutions when a is greater than 0.

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.From m square + 1 = 5, the bridge row is dug and m = 2

And because of the intersection of two points, so =4(m+1)squared-4(msquared+1)=8m>0 so m=2, blanking analytic formula y=x2-6x+5

So the ab points are (5,0)(1,0).

2.Similar when ob oc=op oq or ob oc=oq op (p<5) ob oc=1 5

op=5-t，oq=t

i.e. 1 5 = (5-t) t

Get t = 25 6

When ob oc=oq op.

1/5=t/（5-t)

t=5 6, so t1=25 6, t2=5 6

Answer: The 1 function intersects the y-axis at point c(0, 5).

m 2 + 1 = 5 so m = 2

The function has two real roots, so with friends >0, b 2-4ac>0, that is, [2(m+1)] 2-4(m 2+1)>0

Simplifying 4m 2+8m+4-4m 2-4>0 to get m>0 m can only take the state brother value +2

The analytic formula of the function is y=x 2-6x+5

It is easy to obtain the coordinates of point b as (1,0) and the coordinates of point A as (5,0)Solution: 2 should make obc similar to opq.

As long as the oq:op=1:5 or oq:op=5 is satisfied, the condition is satisfied.

oq=top=5-t

t:5-t=1:5, t=5 Book 6

t:5-t=5, t=25 6 certificate completed.