A high school math problem about the range of functions

Not =,= commutation.

x-1=t》0

t^2+1=x\

y=2t^2+2-t t>=0

2(t-1/2)^2-1/2

zhiyv=(-1 2, positive infinity) includes -1 2

Mine: It can be understood that way;

Solution: If y=2x- (x-1) is meaningful, then x-1 0, i.e. x 1;

When x=1, y=2 (maximum);

When x>>1, the value of y tends to infinity (2x> (x-1);

So the range is y [2, ].

Done, look!

Solution:Find the range of f(x) firstIt is easy to see from the diagram below that on the x-interval [-1, 2].The range of f(x)=x -2x is [-1, 3].

f(x)=x -2x.

Then find the range of g(x).Because a>0, g(x)=ax+2 increases monotonically over the x-interval [-1, 2], soIts value range is: -a+2 g(x) 2a+2

And since on the interval x [-1, 2], f(x)=g(x), the range of f(x) [-1, 3] should be the subset of the range of g(x).

So there is: -a+2 -1, 2a+2 and 3

Find: a 3That is, the value range of the real number a is: [3, +

Solution: when x1 [1,2], the range of y x 2-2x y [l,3];

x2 [-1,2] allows the value of y ax 2, a 0 in the range y [ a 2, 2a 2]. Shengshen changed the joke to f(xl) g(x2), so a 2 l, 2a 2 3, and a 3.

When x1 [ 1,2], the value range of y x1 2-2x1=(x1-1) 2-1 is a=[ l,3]; Friendly faction.

x2 [-1,2], y ax2 2(a 0) is the increasing function, and its range b = [ a 2, 2a 2].

Since there is always x2 [-1,2] for any x1 [ 1,2], such that f(xl) g(x2), then a is a subset of b, so a 2 l, and 2a is good for Nahe3, and the solution is a 3, which is what is sought.

The function y=g(x) is the subtraction function defined on r, the range is (c,d) the function y=g(x) is the increase function defined on r, the range is (-d,-c) and the function y=f(x) is the increase function defined on r, the range is (a,b) the function y=f(x)-g(x) is the increase function defined on r, and the range is (a-d,b-c).

The function y=g(x) is the subtracting function defined on r, and the range is (c,d) further, and the function y=-g(x) is the increasing function defined on r, and the range is (-d,-c).

The function y=f(x) is an increasing function defined on r, the range is (a,b), and the function y=f(x)-g(x) is an increasing function defined on r, and the range is (a-d,b-c).

Your tip is problematic.

Since the slope from the moving point on y=x to (1, 1) is in the open range (1,1), in fact, the answer to this question should be easy to determine. Obviously, y=(x+1) (x-1)=1+2 (x-1) i.e. y-1=2 (x-1) Obviously, it is an inverse proportional function, and the image is a hyperbola. The asymptote is x=1 and y=1 and can be obtained by translating the image y=2 x one unit to the right and then one unit upward.

The original hyperbola had a range of y that was not equal to 0, and now the range is any real number that is not equal to 1.

We have learned important inequalities: (1) when a,b>0, (a+b) 2>= ab, (2) when a,b,c>0, (a+b+c) 3>= abc under the cubic root sign

The above and the second question use the formula (2).

y=x 2(1-x), because there is a condition 00, 1-x>0y=x*x*(1-x) [If you add up with a plus sign in the equation, x can't be eliminated, you can only multiply a 2 here at (1-x), and multiply it by a 1 2] = 1 2*x*x*(2-2x)<=1 23rd power = 1 2 (2 3) 3rd power = 4 27

The equal sign holds if and only if x=2-2x, so x=2 3

The preceding brackets cancel each other out.

The value range of the function can be obtained from the image of the function by translating the image of the inverse proportional function y=2 x to the right by 1 unit, and the value range of the function is (- 0) (1 2,2).As shown in Fig.

f（x）=ax+b/(x^+1)=y

yx^2-ax+y-b=0

Discriminant: a 2-4y(y-b)>=0

y^2-yb-a^2/4<=0

4, -1 is the equation y 2-yb-a 2 4 = 0 two roots touch the god Changshen.

b=4-1,-a^2/4=4*(-1)

b = 3, a = 4 or a = -4

a+b=7 or a+b=-1

1. (x)=｛x-3（x≥10）

f(x+5)(x＜10)

1) Find f(4) 4<10

So f(4) = f(4+5) = f(9).

So f(9) = f(9+5) = f(14).

So f(14)=14-3=11

i.e. f(4) = 11

2) Find f[f(5)].

Similarly, f(5)=f(10)=10-3=7f[f(5)]=f(7)=f(12)=12-3=92f(x)={x +2(x less than or equal to 2)2x (x greater than 2).

1) Find f(-4).

4<2 Use f(x)=x +2

f(-4)=(-4)²+2=18

2) If f(x0)=8, find x0

There are two scenarios.

First: if x0>2 f(x0)=2x0=8 the solution is x0=4 Second: if x0 2 f(x0)=x0 +2=8x0 =6

The solution yields x0 = 6>2 (rounded) or x0 = - 6

In summary: x0 = 4 or - 6

To solve the problem of piecewise functions, it is important to note that the expressions of different intervals are also different.

f(x+5)(x＜10)

1）f(4)=f(9)=f(14)=14-3=11.

2）f(5)=f(10)=10-3=7,f[f(5)]=f(7)=f(12)=12-3=9.

Less than or equal to 2).

2x (x is greater than 2).

1)f(-4)=(-4)^2+2=16+2=18.

2) When x0<=2, f(x0)=x0 2+2=8, x0 2=6, x0=- 6;

f(x0)=2x0=8 when x0>2, x0=4

f(x=4<10)=f(x=4+5=9<10)=f(x=9+5=14>=10)=14-3=11;

f(x=5<10)=f(x=5+5=10=10)=10-3=7;

f[f(5)]=f(7)=f(12)=12-3;

f(-4)=(-4)^2+2=18;

If x0>2, then 2x=8, push x=4;

If x0<=2, then x 2+2=8, push out x=- root number 6

From the meaning of the title:

mx +4x+m+2 0 and x -mx+1≠0 is constant on x r m 0 (opening up), 1 0 (no intersection with x-axis), 2 0 (similarly, this opening is already up).

Solution 5 - 1 m 2

The general way to find the domain of a function is to change y=f(x) to x=f(y), and then evaluate the domain according to the definition domain of the function.