The image of function f (x) in interval [a, b] is a continuous curve. Why is it continuous

The image of function f (x) in interval [a, b] is a continuous curve. Why is it continuous

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The basic elementary function and elementary function (the function obtained by the basic elementary function through four finite operations and coincidence) are continuous in their definition domain
The above theorem shows that the elementary function is a continuous function in the definition domain, so the image is a continuous curve

The function f (x) = (A-1 / 2) e ^ 2x + X (a ∈ R) is known. If on the interval (0, +), the image of function f (x) is always below the curve y = 2ae ^ X Find the value range of A. there is a great reward for answering in time

The function f (x) = (A-1 / 2) e ^ 2x + X (a ∈ R) is known. If the image of function f (x) is always below the curve y = 2ae ^ x on the interval (0, +), find the value range of A
Analysis: ∵ function f (x) = (A-1 / 2) e ^ 2x + X (a ∈ R). On the interval (0, + ∞), the image of function f (x) is always below the curve y = 2ae ^ X
That is, 2ae ^ x-f (x) > 0 is always true
Let g (x) = 2ae ^ X - (A-1 / 2) e ^ (2x) - X
When a = 0
g(x)=1/2e^(2x)-x
Let g '(x) = e ^ (2x) - 1 = 0 = = > x = 0
G '' (x) = 2E ^ (2x) > 0, ‡ g (x) takes the minimum value 1 / 2 > 0 at x = 0
‡ meet the requirements of the topic;
When a < 0
g(x)=2ae^x-(a-1/2)e^(2x)-x
Let g '(x) = 2ae ^ X-2 (A-1 / 2) e ^ (2x) - 1 = 0 = = > e ^ x = 1 or e ^ x = 1 / (2a-1)
∴x1=0,x2=-ln(2a-1)
g’’(x)=2ae^x-4(a-1/2)e^(2x)==> g’’(0)=2-2a
g’’(-ln(2a-1))=2a/(2a-1)-4(a-1/2)/(2a-1)^2=(4a^2-6a+2)/(2a-1)^2
∵a<0
‡ G '' (0) > 0, G (x) takes the minimum value at x = 0, G (0) = a + 1 / 2
‡ when - 1 / 2, when a > 0
g(x)=2ae^x-(a-1/2)e^(2x)-x
Let g '(x) = 2ae ^ X-2 (A-1 / 2) e ^ (2x) - 1 = 0 = = > e ^ x = 1 or e ^ x = 1 / (2a-1)
∴x1=0,x2=-ln(2a-1)
g’’(x)=2ae^x-4(a-1/2)e^(2x)==> g’’(0)=2-2a,g’’(-ln(2a-1))=2a/(2a-1)-4(a-1/2)/(2a-1)^2=(4a^2-6a+2)/(2a-1)^2
∵a>0
‡ 00, G (x) takes the minimum value at x = 0, G (0) = a + 1 / 2
When 1 / 2A = 1, G (x) is reduced in the definition field;
A> 1, G '' (0) < 0, G (x) takes the maximum value at x = 0; G '' (- ln (2a-1)) > 0, G (x) takes the minimum value at x = - ln (2a-1);
‡ 0 to sum up: when - 1 / 2

If the domain of function f (x-1) is closed interval 1 to 2, then the domain of F (x) is

∵1≤x≤2;
∴0≤x-1≤1;
The definition field of F (x) is [0,1]
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Function y = - 2x2 + 4 ｜ x ｜ + 1, find the image, definition field, value field and monotone interval of the function

The image of the function is omitted
The definition domain is r
The value range is (negative infinity, 3]
The monotonic increasing interval is (server large, - 1] and [0,1]
The subtraction interval is [- 1,0] and [1, positive infinity)

Find the value range of x ^ 2-4x + 5 under the function y = root sign

Greater than or equal to 5 or less than or equal to - 1

Function y = The range of 5 + 4x − X2 is __

Let t = 5 + 4x-x2, from the image and properties of the quadratic function, we can get that the maximum value of the function is 9
To make the analytic expression of the function meaningful, t ≥ 0
Therefore, 0 ≤ 5 + 4x-x2 ≤ 9,
So 0 ≤
5+4x−x2≤3
So the function y =
The range of 5 + 4x − X2 is [0, 3]
So the answer is: [0, 3]