How to solve complex function problems in high school 10

It's just two of the same formula, and to solve two different values, look at the figure. Defines the two endpoint values for the domain.

First of all, the function is monotonic, and when c>1, the single increases. The halving means that there are 2 intersections between the function and the straight line y=x 2 (or y=-1 2(x-a-b), which is not considered for the time being).

A function problem in high school, how to solve it, thank you for the writing process.

lgx=8-x，10ʸ=8-y

It is equivalent to x is the root of lgx=8-x, that is, the abscissa of the intersection of y=lgx and y=8-x images, and y is the root of 10=8-x, i.e., the abscissa of the intersection of y=10 and y=8-x.

Because the image of y=lgx and y=10 is symmetrical with respect to y=x, its intersection with the line y=8-x is also symmetrical with respect to the line y=x. So this translates to y=8-x and x, the intersection of the y-axis is symmetrical with respect to y=x. Since y=8-x and x,the intersection of the y-axis is (8,0) and (0,8), their midpoint is (4,4).

So (x+y) 2=4

i.e. x+y=8

The axis of symmetry of the function f(x) is x=a, and considering several divisions of a, it is actually considering the position of the axis of symmetry The axis of symmetry is on the left side of the interval, then f(x) is monotonically increasing on [-2,2] The axis of symmetry is on the right side of the interval, then f(x) is monotonically decreasing on [-2,2] The axis of symmetry is between the interval [-2,2], then only the minimum value of the function can be determined, that is, f(x) has a minimum value when x=a, but its maximum value cannot be determined, because the left side of the symmetry axis decreases, Incrementing on the right, you can't determine the size of f(-2) and f(2), i.e. you can't determine the maximum value of the function.

Therefore, it is necessary to subdivide it further, the middle point of -2 and 2 is 0, then it is divided into (-2, 0) and [0, 2) two ends, so that the symmetry axis is in a certain interval according to the symmetry to know the size of f(-2) and f(2), for example, when the symmetry axis is in (-2, 0), then x=2 is far away from the symmetry axis, at this time it is f(2) f(-2).

One: If the axis of symmetry x=a does not fall on [

interval, the function is monotonic, but when the axis of symmetry is on the left of a given interval, it is monotonically increasing, the minimum is x=-2, and the maximum is taken at x=2; When the axis of symmetry is on the right side of a given interval, the function is monotonically decreasing, with the maximum at x=-2 and the minimum at x=2.

2: If the axis of symmetry x=a falls on [

interval, then the minimum value of the function must be taken when x=a, but where to take the maximum, it will be divided into two types of cases When a is from -2 to 0, then 2 is farther away from a than -2 from a, and the maximum value of the function is taken at x=2; When a is from 0 to 2, -2 is farther away from a than 2 from a, and the maximum value of the function is taken at x=-2.

To sum up, A should be considered in four categories.

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.

g(x)=f(x-1)odd function: g(x)=-g(-x)f(x-1)=-f(-x-1)=-f(x+1), so f(x)+f(x+2)=0

f(x+2)+f(x+4)=0②

=f(x)-f(x+4)=0,f(x)=f(x+4) is the original formula s for the periodic function, 2s=f(1) f(2)+f(2010)+f(2011) =2f(2)=-4,s=-2

f(x)=f(-x)

f(x-1)=f(-x-1);

This is the conclusion based on the conditions of the problem, and then the first formula of the x-1 generation is used to obtain f(x-1)=f(1-x). So f(1-x)=-f(-x-1) is reversed every two unit functions, and the value of the two unit functions is reversed. So f(x) takes 4 as a period.

g(x)=f(x-1) odd function. g(x)=-g(-x), i.e. f(x-1)=-f(-x-1)=-f(x+1), so f(x)+f(x+2)=0

Original formula = f(1) + f(2) = -2

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.

Swap x with 1x.

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

There is another 2f(x)+f(1 x)=3x

Solve a system of binary linear equations.

f(x)=2x-1 x

In fact, if you look at my answer again, f(x)=1-1 (a x+1), then you can know that when x takes negative infinity and positive infinity respectively, f(x) obtains the minimum and maximum values 0 and 1 respectively, and f(0)=1 2, you can know that when x>0 , 1 20, [f(x)-1 2]=0, [f(-x)-1 2]=-1, x<0 is the same, so the range is actually the last step is written wrong, look forward and the range is.

Change the line, f(x)=1-1 (a x+1), then we can know that when x takes negative infinity and positive infinity respectively, f(x) obtains the minimum and maximum values of 0 and 1 respectively, and f(0)=1 2, we can know that when x>0 , 1 20, [f(x)-1 2]=0, [f(-x)-1 2]=-1, x<0 is the same, so the value range is.

The key is that when a>1 or 01, f1(x) passes (0,1) and increases the function; f2(x) over (0,0), increasing function, concave; f3(x) over (1,0), the increment function, located to the right of x 1!

a<1, f1(x) over (0,1), subtract the function; f2(x) over (0,0), increasing function, convex; f3(x) over (1,0), a subtractive function, between x 0 and x 1!

Select BI'll talk about one, for example, the point that a term is over (0,1) is obviously y=a to the x power, and the graph knows a>1, and the remaining graph that can't be logx can only be a power of x, but it should be an increasing function. Therefore, item A is incorrect.

When 01, the derivatives of all three functions increase in the first quadrant;

Thus, for the same a, the monotonicity of the three functions is the same.

In this way, it is not difficult to see that it is B.