Function value range, how to find the value range in the function

f(x)=x^2+2x+a/x x∈[1,＋∞

Derivation f'(x)=2x-a/x^2+2=（2x^3+2x^2-a）/x^2 x∈[1,＋∞

Now with f'(x) Discuss the image of f(x):

Let g(x)=2x 3+2x 2-a know x [1, i.e., x>0 is constant, so f'The plus or minus of (x) is the same as g(x) g'(x)=6x 2+4x x>0 g'(x) Evergrande at zero, g(x) at [1, constant increase on the top, i.e. f'(x) In [1, on constant increase.

f'(1)=4-a

1. When f'(1) >0 (i.e., a<4) and because of f'(x) in [1, on constant increase, so f'(x) at [1, Evergrande at zero, i.e. f(x) constant increase. To make x [1, f(x)>0 constant, just let f(1)=3+a>0 in this case a (-3,4).

2. When f'(1) When <0 (i.e., a>4), f(x) decreases first and then increases, f'(x.)=0 x.is the minimum point of f(x).

by f'(x.) = 0 to get 2x^3+2x.2=a (at this time x>1 and a<4) because y=2x 3+2x 2 is constantly increasing on r.

x>1 2x 3+2x 2>4 so there is no solution at this time.

In summary, a (-3,4).

Functionalize as f(x)=[(x+1) 2+a-1] xthen [(x+1) 2+a-1] x>0

Since x [1,+ so (x+1) 2 is evergreen in 4, then only a-1>-4 is sufficient.

then a>-3

The value of a can be (-3,+.)

It is evident from the observation that when a>=0, for all x>=1f(x) = x 2 + 2x + a x is the sum of three positive numbers and 0 is 0 for a<=0, -a x is the increasing function, and f(x) = x 2 + 2x + a x is the sum of the three increasing functions is also an increasing function. The minimum value of f(x) is f(1) = 3 + a > 0

A > 3 is available

On the whole, a 3,+

The value range of the function is mainly paid attention to the following points:

1. Practical questions should be considered.

2. The denominator is not 0

3. The true number of logarithms should be greater than 0

4. Negative numbers cannot be opened to the even square root.

5. Trigonometric inverse trigonometric functions should conform to their defined domains.

6. At the same time, there are radical, fractional, logarithmic, and trigonometric inverse trigonometric functions that should conform to their defined domains and take the intersection of each defined domain.

The principle of finding the range of values of the independent variables of a function is:

1) The analytic formula is an integer, and the independent variable can be taken as a real number.

2) The analytic formula is a fraction, and the value of the independent variable should be such that the denominator is not equal to zero.

3) The analytic formula is irrational, if it is a quadratic radical, the value range of the independent variable should be greater than or equal to zero, and if it is a cubic radical, the independent variable can be taken as a real number.

4) If the analytic formula is a combination of the above forms, the value range of the independent variables satisfies their respective conditions at the same time.

The argument is generally a guarantee that the function is meaningful. (e.g. y=lnx,x>0.)

y=ln(lnx), lnx>0 and x>0; i.e. x>1)

The dependent variable is determined by the range of independent variables. It involves the increase or decrease of the function, and finds the range of the value of the function within the range of the corresponding independent variable, that is, the range of the value of the dependent variable!

The inequality is deformed as [(2t-1)x-12t] (2x-12) 0 two zeros: 6 and 12t (2t-1).

Since t>1 2, it is easy to get 12t (2t-1)>6, so the x range is 6

Search: How to find the value range.

Several common methods for function ranges.

1 Direct method: use the range of common functions to find it.

The domain of the primary function y=ax+b(a 0) is r, and the value range is Sun Mingqiaor;

The inverse proportional function is defined in the domain and the value range is;

The quadratic function is defined in the domain r, and when a>0 is the key, and the range is ; When a0, when x0, then when , its minimum value;

a0) or the maximum value (a

It depends on the range of values of the independent variable and the actual meaning of the function.

Look at the function relation, denominator, open even power, power zero, the base of the logarithmic function, and the requirements of the practical problem.

Find the range of the function.

Please take a look at the original question.

3-k＜0

k＜0k＞3k＞0

k 3 friends, please [adopt the answer], your adoption is the motivation for me to answer the question, thank you.

How to find the value range of the function:

1. For functions that are very familiar with increase and decrease, you can see the range of values by drawing an image directly.

2. For the function that is not familiar with the increase and decrease, you can use the derivative, find the station, draw the image, and finally find the extreme value, compare the extreme value, and obtain the maximum value, and finally find the value range.

Derivative is an important fundamental concept in calculus. When the independent variable x of the function y=f(x) produces an incremental δx at a point x0, the ratio of the incremental δy of the output value of the function to the incremental δx of the independent variable is at the limit a when δx approaches 0 if it exists, a is the derivative at x0 and is denoted as f'(x0) or df dx (x0).

Method 1: Discuss from x to y;

Method 2: Use y to represent x, and use the value range of x to obtain the inequality group to find the value range of y to see the process comprehension.

If you are satisfied, please adopt it in time. Thank you!

It is known from the meaning of the title.

y=（2x-1）/（x-1）=〔2(x-1)+1〕/(x-1)=2+1/(x-1)

Since y=x is an increasing function, when x r, y=x-1 is an increasing function, when x r, y=1 x is a decreasing function, when x (0,+ then the function y=2+1 (x-1), when x [0,1), the function is a subtraction function, and when x (1,3], the function is a decreasing function.

When x=0, y=1, and when x=3, y=

So when x [0,1), the function value y (-1], when x (1,3], the function value y [,

y＝(2x-1)/(x-1)

2x-2+1)/(x-1)

2+1/(x-1)

Given that the range of x is 0 x 3 and x ≠ 1, it can be seen that on [0,1), y increases with the increase of x and decreases with the increase of x at (0,3].

When x 0, y 1, x tends to 0, y tends to infinity, x 2, y 3, x 3, y 5 2, so the value range of the function value is (1, +

Because f'(x)=-3x 2+2ax-1 , and since the shed fiber function is monotonic on r, f'(x) On r, the sign constant orange is negative for chain slag, then discriminant = 4a 2-12< = 0 , the solution is - 3< = a< = 3.

Method 1:

y＝［（x－1/2）＋2a＋1/2］/（2x－1）＝1/2＋（2a＋1/2）/（2x－1）。

In the interval (1 2, on, 2x 1 0, and increases as x increases.

To make y an increasing function in this interval, we need 2a 1 2 0, 2a 1 2, a 1 4.

The value range of a that satisfies the condition is (1 to 4).

Method 2: y (x 2a) (2x 1), y 2x 1) 2 (x 2a) (2x 1) 2 (2a 1) (2x 1) 2.

Let y 0, get: (4a 1) (2x 1) 2 0, 4a 1 0, 4a 1, a 1 4.

The value range of a that satisfies the condition is (1 to 4).

1。If you learned the derivative, do this: y'=(-1-4a) (2x-1) 2>0 is constant on (1 2, infinity), so a<-1 4

2。If you don't learn the derivative, then do this: y=1 2+(2a+1 2) 2x-1, then 2a+1 2<0, so a<-1 4

After the separation constant, the molecule is less than 0

The argument is generally a guarantee that the function is meaningful. (e.g. y=lnx,x>0.) y=ln(lnx), lnx>0 and x>0; i.e. x>1)

The dependent variable is determined by the range of independent variables. It involves the increase and decrease of the function, and finds the range of the value of the Shengzikai function within the range of the value of the corresponding independent variable, that is, the range of the value of the dependent variable!

Independent variables The denominator of the fraction is not zero The number of open squares of the root formula is greater than or equal to the chain hole and is 0 The logarithm The true number is greater than zero In real life, the shed and the meaningful value range can be defined and determined.

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