A simple high school math problem that I haven t figured out, please help me, math enthusiasts

f(0+0)=f(0)+f(0)

f(0)=2f(0)

f(0)=0 Got it here?

As for this question for you.

If f(x)=x+1 f(y)=y+1, then f(0+0)=f(0)+f(0)=2 instead of equal to 0, which is contrary to the problem-solving process."

When f(x)=x+1 f(y)=y+1, f(0+0)=f(0)+f(0)=2 is not true at all.

In fact, the example you gave f(x) does not conform to f(x+y)=f(x)+f(y).

f(o)=0 is not necessarily an odd function to verify.

f(0)=0.

Then f(0)=f(x)+f(-x), i.e., f(x)+f(-x)=0, is an odd function, which is proved by definition.

f(0)=2f(0) ∴f(0)

As for the example you gave, although f(x) is an arbitrary function, if you want to overturn it in the other way, the example has to be an odd function and must make sense at x=0, but obviously f(x)=x+1 is not an odd function.

The piecewise function, when x 1, is obviously a one-time function, and the subtracting function is only a one-time term coefficient (3a-1) 0, i.e., a 1 3;

When x 1, f(x)=log takes a as the base of the logarithm of x, minus 0 a 1, and because at this time f(x) f(1)=(3a-1) 1+4a=7a-1, log takes a as the base 1 logarithm is 0, so 0 7a-1, a 1 7,;

In summary, 1 7 a 1 3

(3a-1)x+4aax-small is equal to 1, subtraction, (3a-1)<0, a<1 3

logx loga x is greater than 1, minus the function, 0=log1 loga=0, a>=1 7

So 1 7<=a<1 3

Because it is a subtraction function on r, it is also a subtraction function on x small equal to 1 3a-1<0 so a is less than 1 3

But x=1 is f(x)=7a-1 and is also a subtractive function on x greater than 1, so logx loga x is less than 0

Substitute x=1 into logx loga x logx loga x=0 so 7a-1 "0 so 1 7 small equals a

The sine value of the inclination angle of the line l is 3 5

k = tana = 3 4 or k = -3 4

Let the equation of the straight line be y=3x 4+b1 or y=-3x 4+b2x=0 and y=b1

y=0,x=-4b1/3

2b1²/3|=6

b1=±3x=0,y=b2

y=0,x=4b2/3

2b2²/3|=6

b2 = 3 is the equation for the straight line.

y=3x/4+3

or y=3x 4-3

or y=3x 4-3

or y=3x 4-3

Because set b is not an empty set, there is a solution to x in set b, and because set b is a quadratic equation, there are two cases, there is one x solution or there are two x solutions.

Moreover, set b is contained in set a, which can be a column of equations.

When b has one element, x=-1 or 1

and b 2-4ac = 0, i.e. 4a 2-4b = 0a 2 = b when x = -1, 2a + b + 1 = 0

The two equations are combined, and a=-1 and b=1 are obtained

x=-1, use the same method.

Get a=1, b=1

When b has two elements, i.e., x has two solutions, substituting x=1 or x=-1 into a quadratic equation yields: 1-2a+b=0

1+2a+b=0

Get a=0, b=-1

1) a=2)0 is not, if 0 is, then (1+0) (1-0)=1 is also, which does not fit the topic. a=5a=

3) If a≠0, a≠1 belongs to a, a=

Well, the first one.

The first and second questions are simple calculations, and the third question is just to find the law, if you want to prove it, it is also simple, bring the letter A in, get the next result, and then bring it into the original formula, until the fifth result is found to be still A, which can be proved by induction.

What you don't understand is why the elements of this set are finite, this is because the elements of the set are different, when the fifth result appears the same as the first result, because the recursive rule does not change, the subsequent result must be the first four in a continuous cycle, then of course there are only four elements in the set.

f(x) is a primary function.

Let f(x)=ax+b

and g(x)=1 4(3+x(2)).

then g(f(x))=1 4[3+(ax+b)(2)]=3 4+a(2) 4*x(2)+1 2abx+1 4b(2).

and g(f(x))=x(2)+x+1

b = 1, a = 2 or b = -1, a = -2

i.e. f(x) = 2x+1 or f(x) = -2x-1

bb'//aa'So pm ma=pb aa'

bb'//cc'So pn nc = pb cc'

aa'=cc"So pm ma=pn nc is proven.

It's inconvenient, hehe, I don't know if I understand.

Connect A'c', because m n is the two midpoints, so mn is parallel a'c'。Because a'c'In plane a'b'c'd', and a'b'c'd'Parallel to the planar ABCD, so a'c'Parallel ABCD, so MN parallel ABCD

...Upstairs, it can't be proven that MN is the midpoint.

And most likely, it's not the midpoint.

Let a (1 3) = x, b (1 3) = y Simplify a-3a (2 3)+5a (1 3)=1x 3-3x 2+5x-1=0

x(x-1)(x-2)+3(x-1)=-2(x-1)((x-1)^2+2)=-2

x-1)^3+2(x-1)=-2

b-3b^(2/3)+5b^(1/3)=5y^3-3y^2+5y-5=0

y(y-1)(y-2)+3(y-1)=2

y-1)^3+2(y-1)=2

Function f(x')=x'^3+2x'=x'(x'^2+2)f(-x')=-f(x')

f(x') is an odd function with origin symmetry.

The above system of equations can be seen as f(x')=2 f(x')=-2, i.e. y-1=x0

x-1=-x0

x-1+y-1=x0+(-x0)=0

i.e. x+y=2

Because a-3a (2 3) + 5a (1 3) = 1 order a (1 3) = t

Get: t 3-3t 2+t = 1

T = 2 is obtained because b-3b (2 3) + 5b (1 3) = 5 let b (1 3) = f

Get f 3-3f 2 + 5f = 5

Find: f and then t+f= is the answer.