Let min {a, B} denote the minimum value of a and B. Let f (x) = min {x + 2,10-x}, what is the maximum value of F (x) Why is there only the lower part of the image? Isn't the value range just x > = 0?

Let min {a, B} denote the minimum value of a and B. Let f (x) = min {x + 2,10-x}, what is the maximum value of F (x) Why is there only the lower part of the image? Isn't the value range just x > = 0?

The simultaneous solution of y = x + 2 and y = 10-x gives x = 4 and y = 6
That is, when x ≤ 4, the value of function y = x + 2 is less than or equal to y = 10-x
∴f(x)=min{x+2,10-x}=x+2.
When x > 4, the value of y = 10-x is less than that of y = x + 2
∴f(x)=min{x+2,10-x}=10-x
Therefore, the maximum value of function f (x) is obtained when x = 4
The maximum value is 6

Min {a, B, C} is used to represent the minimum value of a, B, C, min {0,2,3} = 0; if {- 1,2, a} = a, what is the value range of a
It's better to make the process clear.

a

Let min {x, y} denote the minimum value of X and y, for example, min {0, 2} = 0, min {12, 8} = 8, then the function y = min {2x, x + 2} of X can be expressed as______ ．

When 2x ≥ x + 2, that is, X ≥ 2, the function y = min {2x, x + 2} = x + 2; when 2x < x + 2, that is, x < 2, the function y = min {2x, x + 2} = 2x. So the answer is x + 2 (x ≥ 2) 2x (x < 2)

Let min {x, y} denote the minimum value of X and y, for example, min {0, 2} = 0, min {12, 8} = 8, then the function y = min {2x, x + 2} of X can be expressed as ()
A. y=2x(x＜2)x+2(x≥2)B. y=x+2(x＜2)2x(x≥2)C. y=2xD. y=x+2

It is known that when the value range of X is not given, the size of 2x and X + 2 can not be determined, so it can not be directly expressed as C: y = 2x, D: y = x + 2. When x < 2, we can get x + x < x + 2, that is, 2x < x + 2, which can be expressed as y = 2x. When x ≥ 2, we can get x + X ≥ x + 2, that is, 2x ≥ x + 2, which can be expressed as y = x + 2

The cube of 2x - 14 = 0,

2x^3-14=0
2x·x^2-7=0
x^3-7=0
x^3=7
x=3√7
You know, it's very detailed! √ root

LIM (x tends to infinity) xsin (3x / 1 + x ^ 2)=

Let y = 3x / (1 + X & # 178;)
x->∞,y->0
So the circular limit is equal to the limit of X * 3x / (1 + X & # 178;) = the limit of 3 / (1 + 1 / X & # 178;)
= 3

lim(3x+7/3x+5)∧6x+1…… X approaches infinity

LIM (1 + x ^ 2) ^ (1 / LNX) x tends to be positive infinity

=e^ lim ln(1+x^2) / lnx
=e^ lim (2x/(1+x^2) ) / (1/x)
=e^ lim (2x²/(1+x^2) )
=e^ lim (2/(1/x² +1) )
=e^2

How to do LIM (lnx-x / E) x - > infinity

The derivative of LNX is 1 / x, when x is very large, LNX has been basically unchanged. But the derivative of X / E is always 1. The larger x is, the larger X / E is. So LNX is much smaller than X / e. the above is a perceptual explanation. Mathematical derivation can be done as follows: lnx-x / E = ln [x / exp (x / E)], assuming y = x / E, Y - > infinity, the lion above is equal to L

lim(lnx)-1/(x-e)

lim(x→e) (lnx-1)/(x-e)
=lim(x→e) (1/x)/1
=lim(x→e) 1/x
=1/e