It is said that 99 people can t answer the super difficult high school function math problem

f(5-x 2)=(5-x 2) 2+2(5-x 2)-1=g(x) The derivation of this function yields: g'(x)=2(5-x 2)(-2x)-4x=4x(x 2-6)=4x(x+6 (1 2))(x-6 (1 2)).

Discussion: In 4 contiguous intervals:

1.(-infinity, -6 (1 2)], g'(x) <0, the function is monotonically decreasing.

g'(x) = 0 minimum.

g'(x) >0, the function increases monotonically.

g'(x) = 0 maximum.

g'(x) <0, the function is monotonically decreasing.

g'(x) = 0 minimum.

7.(6 (1 2), positive infinity], g'(x) >0, the function increases monotonically.

Solution: Let 5-x 2=t

Then f(t)=-t 2+2t-1

x^4+8x^2-16

f(t)=-4x^3+16x

4x(x+2)(x-2)

Let f(t)=0

then x=0, x=2, x=-2

It is made up of the number line root method.

When x belongs to (-infinity, -2), f

t) >0, the function increases monotonically.

When x belongs to (-2,0), f

t）<0

When x belongs to (,f

t)>0...

When x belongs to (2, positive infinity), f

t）<0...

1. For the content in the textbook, it is best to preview it before class, and the targeted practice questions after class must be done carefully, not lazy, and you can also repeat the class example problems several times when reviewing after class, after all, take good class notes when you are in class. "A good memory is better than a pen". For the solution of mathematical, physical and chemical problems, it is not enough to rely on the general idea in the head, but only after careful pen calculation can we find the difficulties and master the solution method, and finally get the correct calculation results.

2. Secondly, we should be good at summarizing and classifying, looking for commonalities and connections between different question types and different knowledge points, and systematizing the knowledge we have learned. To give a specific example: in the function part of higher algebras, we learned several different types of functions, such as exponential functions, logarithmic functions, power functions, trigonometric functions, etc.

But if you compare them and summarize them, you will find that whatever kind of function you need to grasp is its expression, image shape, parity, increase and decrease, and symmetry. Then you can make the above contents of these functions in a big **, and compare them to understand and remember. When solving problems, pay attention to the combination of function expressions and graphs, and you will definitely get much better results.

3. Finally, it is necessary to strengthen after-class practice, in addition to homework, find a good reference book, and try to do as many practice questions as possible in the book (especially comprehensive questions and application questions). Practice makes perfect, so that you can consolidate the effect of classroom learning and make your problem solving faster and faster.

The function f(x) is an increasing function that defines the domain over (0, positive infinity), and f(x y) = f(x)-f(y), f(x) = loga x , a>0 and not equal to 11, find the value of f(1) f(1) = 0

2. If f(6)=1, the solution inequality f(x+3)-f(1 x) 2a=6 , f(x)=log6 x

log6(x+3)-log6(1/x)<2log6(x^2+3x)<2

log6(x^2+3x)0

3/2-√153/2x<√153/2-3/2

1.Let x=y=1, then f(1)=f(1)-f(1)=0;

2。According to f(x y) = f(x)-f(y), f(x+3)-f(1 x) 2=2f(6), it is reduced to:

f(x+3)-f(1/x)-f(6)-f(6)<0;

f(x(x+3))-f(6)-f(6)<0f(x(x+3)/36)<0=f(1)

The function f(x) is an increasing function that defines the domain on (0, positive infinity), so x(x+3) 36>0 and x(x+3) 36<1, the solution is -(3+sqrt(153)) 2 This type of problem is generally solved by the assignment method, and then the monotonicity and the definition domain in the problem are used.

(1) Substitute x=1 and y=1.

f(x/y)=f(x)-f(y)

then f(1)=f(1)-f(1)=0

2) The original inequality f(x+3)-f(1 x)<2 is equivalent to f((x+3) (1 x))<2

i.e. f(x 2+3x)<2

f(6)=1

f(6)=f(36/6)=f(36)-f(6)=1∴f(36)=2

f(x) is an increment function on (0,+.

When 0 is solved, x

3 + root number 153

Take x=y

6 take x=1, y=x; Then f(1 x) = -f(x), so f(x+3) + f(x)<2, f(6)=1, f(x) increases, so x<6

0=f(1)-f(1)=f(1 1)=f(1), the idea, generally f(0)f(1) This special value is made up by itself.

The second question is actually very simple, when you see this kind of question, you should first test whether the formula given by the question can be used f(x+3)-f(1 x)=f(x 2+3x), which is not used, and the requirement is f(x 2+3x)<2

Then the title says that the function f(x) is an increasing function that defines the domain on (0, positive infinity), but in fact, you need to find when f(y) = 2, for example, when y = m, =2;As long as yf(y) is satisfied, it is less than 2, at this time, x 2+3x, if x y=y, it will produce the effect of 2f(y)=f(x), that is, when x=y*y, 2f(y)=f(x), so when y=6, x=36, f(36)=2f(6)=2x, 2+3x-36<0

The idea is very complete, are you satisfied?

Solution: Let 5-x 2=t

Then f(t)=-t 2+2t-1

x^4+8x^2-16

f'(t)=-4x^3+16x

4x(x+2)(x-2)

Order f'(t)=0

then x=0, x=2, x=-2

It is made up of the number line root method.

When x belongs to (-infinity, -2), f

t) >0, the function increases monotonically.

When x belongs to (-2,0), f

t）<0

When x belongs to (,f

t)>0...

When x belongs to (2, positive infinity), f

t）<0...

Solution: f(x))=2ax 2 + 2x -3-a The axis of symmetry is x=-2 2*2a=-1 2a,1, if -1 2a -1, that is, 0 a 1 2, f(-1) 0, from 2a-2-3-a=a-5 0 to get a 5, from 2a+2-3-a=a-1 0 to get a 1, because 1 a 5 and 0 a 1 2 have no intersection, there is no solution;

2, if -1 -1 2a 1, that is, a 1 2 or a -1 2, when a 1 2, the parabolic opening is upward, to meet f(-1 2a) 0, f(-1) 0 or f(1) 0, since 1 2a-1 a-3-a=-1 2a-3-a 0 is constant, by f(-1) 0 or f(1) 0, a 5 is obtained, so a 5 meets the conditions;

When a -1 2, the parabolic opening is downward, to meet f(-1 2a) 0, f(-1) 0 or f(1) 0, since 1 2a-1 a-3-a=-1 2a-3-a 0 is constant, a 1 is obtained from f(-1) 0 or f(1) 0, and a -1 2 is intersected to obtain a -1 2 is eligible;

3. If -1 2a 1, that is, -1 2 a 0, to satisfy f(-1) 0, f(1) 0, 1 a 5 and -1 2 a 0 have no intersection and no solution;

In summary, A-1, 2, or A5

If you don't understand which step, you can ask (o).

f(x)=2ax²＋2x－3－a。

This problem is that there is x [ 1,1] such that 2ax 2x 3 a=0, i.e. (2x 1)a (2x 3)=0.

1. If 2x 1=0, at this time x= 2 2, the solution A does not exist;

2. If 2x 1≠0, then a= (2x 3) (2x 1). Let 2x 3=t, then x=(1 2)(t 3), after substitution, we get a= 2t (t 6t 7) = 2 [t 7 t 6], where t [ 5, 1], thus ( t) 7 ( t) [ 2 7, 8], thus a ( 3 7) 2] [1,".

f(x)=2ax^2+2x-3-a

x=-1 ,f(-1)=2a-5-a=a-5

x=1 f(1)=2a+2-3-a=a-1

f(x)=0

x1+x2=-2/2a=-1/a

x1x2=(-3-a)/2a

1<-1/2a<1 -1<(-3-a)/2a<1

2<1/a<2 -1/3 <-1/a<1

1<1/a<1/3

a>0.

a>1/2 a>3

a>0.

a<-1/2 a<-1

f(-1)=a-5

f(1)=a-1

a>0 a-5>=0 a-1>=0

a>0 a-5>=0 a-1>=0

So when a>=5, or a<-1, [-1,1] has a zero point.

f(x) = f(x) [g(x)+1] [g(x)-1] where f(x) is the odd function bucket.

g(x)+1]/[g(x)-1] =g(x)+g(x)g(-x)]/g(x)-g(x)g(-x)]

1 + g(-x)] 1 - g(-x)] g(-x)+1] [g(-x)-1], is an odd function.

The multiplication of two odd functions is an even function. , f(x) is an even function.