Senior 2 Mathematics, Basic Properties of Functions, Basic Properties of Functions in High School Ma

f(7-5-x)=f(7-(5+x))=f(12+x) and f(7-5-x)=f(2-x)=f(2+x)x is a period of 10, so the interval (-3,7) is a period, where x=2 is the axis of symmetry.

If it is an odd or even function, there must be f(-1)=0, and about 2 symmetry f(5)=0, which does not match the condition.

So it's a non-odd and non-even function.

Since there is symmetry about x=2 and x=7, there are two roots for every 10 unit intervals, so there are a total of 4*(200+200+1)=802 roots.

f(2-x)=f(2+x)

f(7-x)=f(2-(-5+x))=f(2+(-5+x))=f(-3+x)=f(7+x)

So f(x-3) = f(x+7).

f(x)=f(x-3+3)=f(x+7+3)=f(x+10)

f(x)=f(x+10).

A period function with a period of 10.

f(-x)=f(2-(x+2))=f(2+x+2)=f(x+4)≠f(x+10)

So it's not an even function.

Because only f(1)=f(3)=0 on [0,10].

So f(0) ≠0, so it's not an odd function.

So f(x) is a non-odd and non-even function.

f（3）=f（1）=0，f（11）=f（13）=f（-7）=f（-9）=0

So there are two roots in [0,10], and in a period of 10, then in [0,2005] there are 402.

There were 400 in [-2005,0], so there were 802 roots in [-2005,2005].

First, f(x) is not an odd function because f(0)=0 does not hold, and secondly, f(x) is not an even function, because f(-1)=f(1)=0 does not hold (as a sketch).

Because f(x) = f(4-x) = f(14-x), f(4-x) = f[14-(4-x)] = f(10+x) = f(x), i.e. f(x) period is 10

The sketch shows that the function has 2 roots 1,3 on (-5,5), the interval (-2005,2005) contains 401 such intervals, and the number of roots obtained by using periodicity is 401*2=802.

This kind of problem requires more sketches, and the second is to be proficient in using variable substitution to find out the cycle.

Hope it works for you

For example, f(x)=1 x, the definition domain x is not equal to zero, f(-x)=1 (-x), the definition domain requirement (-x) is not equal to zero, f(x 2+x)=1 (x 2+x), the definition domain requirement (x 2+x) is not equal to zero, for example, in the first question, f(x) defines the domain [-2,4], f(-x) should apply the above definition domain to the (-x) in parentheses, so -2<=(-x)<=4, the definition domain is [-4,2], and the g(x) definition domain is the intersection of the above two definition domains, [-2,2].

(1) Since the domain of y=f(x) is [-2,4], the domain of the function g(x)=f(x)+f(-x) is solved by -2<=-x<=4 and -2<=x<=4.

2) If y=f(x) is defined in the domain [-6,2], by 6》2, then y=f(6) is not intended.

General: 1) It is known that the domain of y=f(x) is [a,b] and the domain of y=f[g(x)] is solved by the inequality a<=g(x)<=b.

2) Knowing that the domain of y=f[g(x)] is [a,b], finding the domain of f(x) is equivalent to finding the range of g(x) in the interval [a,b].

The definition domain is [-2,4], that is, the x in parentheses belongs to [-2,4], then the -x in f(-x) parentheses also belongs to [-2,4], and the definition domain of g(x) is [-2,2].

The brackets in question 2 are not clear, in short, the definition domain of f(x) is [a,b], then the definition domain of f(f(x)) is f(x) belongs to [a,b].

2.If it's root number 6, it exceeds the definition of f(x), so it doesn't make sense to say y.

Functions have four main properties:

Bounded, monotonic, parity, periodicity.

1) The uninterrupted function of the image must be bounded in the closed interval, and sinx and cosx are bounded as a whole.

2) Parity is only discussed for functions defined on symmetric intervals, if f(x)=f(-x), it is an even function, and the image is symmetric with respect to the y-axis; If f(x)=-f(-x), it is an odd function, and the image is symmetrical with respect to the origin.

The properties of high school functions mainly include:

1.Monotonicity of functions.

2.Parity of functions.

3.Periodicity of the function.

4.Symmetry of the function.

There is monotonicity, parity, periodicity.

1) Let's start by looking at the definition domain of the function.

x to be satisfied. x+3)(x-1)>=0

Divide the whole real axis into 3 segments, (-infinity, -3], (3,1), [1,+infinity).

Select any 3 numbers in the 3-segment interval and bring in the inequality test above.

For example, if we bring -4, 0, and 2 to the left of the inequality above, we can see that -4 and 2 satisfy the inequality, while 0 does not satisfy the equal ten.

Next, we look at the endpoints -3 and 1 of the interval, and finally determine the definition domain of the function.

infinity, -3].

And. 1, + infinity).

2) For the function y=f(x)>=0

, sqrt[f(x)] is the square root of f(x). then due to.

sqrt[f(x1)]

sqrt[f(x2)]

f(x1)-f(x2)]/

Therefore, the function y=f(x)>=0 is consistent with the monotonic interval of z=sqrt[f(x)].

Thus, just examine the function z=(x+3)(x-1) in (-infinity, -3] and . 1, + infinity) on monotonicity.

z=(x+3)(x-1) is a parabola with an opening pointing upwards.

Thus, it is monotonically decreasing at the interval (-infinity, -3) and monotonically increasing at (1, +infinity).

So, the title is correct.

That is, the monotonically decreasing interval of the function y = root number (square of x + 2x-3) is (-infinity, -3).

1、 x》a

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

1) a<-1 2 (now the function is defined on x》a, the symmetry axis of the quadratic function is x=-1 2, you can't find the minimum value without discussion, you have to make sure that the minimum value you are looking for can be obtained, it must be meaningful).

b 2a, is it a real number, yes, or that sentence, we can only study the problem in the defined domain, and the discussion of a here is to discuss whether the defined domain contains an axis of symmetry.

fmin=f(-1/2)=3/4-a

2)a》-1/2

fmin=f(a)=a^2+1

2、 xf（x）=a^2-x+a+1

1) a<1/2

fmin=f(a)=a^2+1

2) a》1 2 (This kind of problem has to be combined with graphics, I won't draw pictures) fmin=f(1 2)= 3 4+a

This question is not finished, I still have to sort it out, there is still a lot of work, and you can't bring x in the final discussion, if you can't, I'll make it up for you, I'm going to sleep.

f(x)-2f(1/x)=2/x

f(1/x)-2f(x)=2x

The two equations are solved together.

f(x)=-4x/3-2/3x

y=-f(x)=-(x-1)(x-2)=-x 2+3x-2, the symmetry axis of the function is: x=-(-3) 2=

Because the function is a quadratic function and the opening is downward, so - on the interval (-, the function y=f(x) is a monotonic increase function in the interval [,+, the function y=f(x) is a monotonic subtraction function because the function given in this question is a relatively basic quadratic function, and he is more familiar with his image features, so he can directly judge the increase and decrease according to the image, and what needs to be calculated is only the dividing line of the two monotonic intervals, that is, the axis of symmetry.

The calculated function y=-f(x)=-(x-1)(x-2);

The function has 2 intersections with the x-axis (1,0), (2,0);

The axis of symmetry of the function is x=3 2;

The function opening is facing downward;

The monotonically increasing interval is: negative infinity to 3 2;

The monotonically decreasing interval is: 3 2 to positive infinity.

Axis of symmetry x = 3 2

f(x) in minus infinity to 3 2 single minus.

3 2 to positive infinity single increase.

Then y=-f(x) increases in negative infinity to 3 2 single minus 3 2 to positive infinity.