The process of writing a math function solution in the first year of high school

'=a-1 x 2 because x [1,+infinity]so x 2>0

That is, the minimum value of ax 2-1 0 a 1 x 2 is obtained when x=+ infinity.

so a≤0

Let 1-x=k(k 1).

x=1-k^2

f(k)=1-k^2-k (k≥1）

At (k 1), the range of f(k)max=f(1)=-1f(x) is [-infinity, -1].

3.The axis of symmetry is x=a

When a>1 f(x)max=f(1)=-1+2a+1-a=aa=3, when a<0 f(x)max=f(0)=1-a=3a=-2, when 0 a 1, f(x)max=f(a)=-a 2+2a 2+1-a=a 2-a+1=3

a^2-a-2=0

a=2 or a=-1.

In summary, a=3 or a=-2

Question 1: It is known that the function f(x)=1 x+ax is a monotonic subtraction function on the interval [1,+, and the value range of a is (— 0].

Question 2: The range of 1-x under the function f(x)=x-root number is: (—1]Question 3: The square of the function f(x)=-x +2ax+1-a is taken as the maximum value 3 in the interval [0,1], and the value of a is -2 or 3

The second question above seems to be wrong, e.g. when x=1, f(x)=1

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

So f(x+4).

f[(x+2)+2]

f(x+2)

f(x)f(x+4)=f(x)

t=4 so f(

odd function =-f(

f(so f(

2=1+1=f⑶+f⑶=f⑼

f(a) f(a 1)+2 f((a 1)9) f(x) is an increment function on the defined domain.

a＞(a－1)9

Get a 9 8

a－1＞0a＞1

1＜a＜9/8

f(x) = -2x square + 4

Because the function is even.

so f(x)=f(-x)

2a+ab=0

a = 0 or b = -2

The range of the value range is negative infinity to 4, and if a=0 then the range is negative infinity to 0 or 0 to positive infinity, which does not match the question so b=-2

when x=0 f(x) is maxso f(0)=4

a square = 2so f(x) = 2x square + 4

When b = 0, to satisfy the even function, then a = 0 is required, and the function is constant, and the value range is not satisfied When b is not 0, the function is quadratic and is an even function, then the two symmetries are opposite numbers, and the maximum value of the range is obtained on the y-axis of the symmetry axis, that is, when x=0. The range has a maximum value, so the opening is downward.

Column equations -a-2a b=0

When x=0, 2aa=4

Solve a = root number 2 and b = -2

Write the function yourself.

Analysis: The function f(x)=(x+a)(bx+2a) is an even function f(x)=(x+a)(bx+2a)=bx 2+a(2+b)x+2a 2

Let a(2+b)=0==>a=0 or b=-2

f(x)=bx 2(b≠0) or f(x)=-2x 2+2a2(a r).

f(x) 4).

2a^2=4==>a=±√2

f(x)=-2x^2+4

f(x)=(x +a)*(bx +2a)=bx^2 +(2a +ab)x +2a^2=b[x +(a/b +a/2)]^2 +2a^2 -(a/b +a/2)^2

f(x) is an even function, then: 2a +ab=0

If the f(x) range is on (- 4), then: b<0,2a 2 -(a b +a 2) 2=4

f(x)=bx²+a（2+b）x+2a²

The even function requires f(x)=f(-x), which is actually the axis of symmetry x=-a(2+b) 2b=0

The quadratic function has a range of (- 4), indicating that the function image is open downward, and b<08a b-a (2+b) can be obtained

Cut vertices and sit vertically --=44b, I won't help you solve the best way to solve this kind of function problem is to combine numbers and shapes.

The even function is symmetrical with respect to the y-axis, and since the value range has a maximum value, and it is a bi-function (if b is 0, it cannot be an even function with infinitesimal values), that is, it is n-shaped, that is, the opening is downward, that is, f(0)=2a=4, a=+- root number 2

Then use f(-x)=f(x) to find the b value.

Relationship between the area of a regular triangle and its perimeter:

s = 1/2*sin60°*(l/3)^2 = (√3/36)*l^2

Because: s1 = ( 3 36) * l1 2, s2 = ( 3 36) * l2 2, l1 + l2 = 12cm

s=s1+s2 = ( 2 36)*(l1 2 + l2 2) The minimum value is obtained only when l1 = l2 = 6cm:

s =(√3/36) *36 *2 = 2√3 cm^2

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

Because x>0

So f( x) >2

f(x+△x)-f(x)>0

So it's an increment function.

f(1)+f(1)=f(2)+2

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

f(1)=3

f(a2-2a-2)<3 i.e.

f(a2-2a-2) because of the multiplication function.

So a 2-2a-2<1

1

Because f(x) is an even function.

Therefore f(x) = f(-x).

Because f(x-2) is an odd function.

Therefore f(x-2) = -f(-x-2) (1).

f(x+2)=f(-x-2) (2) (because f(x) is an even function) by (1) (2) has f(x-2)=-f(x+2), so f(x-4)=-f(x) (3).

In (3), replace x with x-4.

f(x-8)=-f(x-4)=f(x)

So f(x) is a periodic function with a period of 8.