Use dichotomy to find a zero point of the function. The reference data are as follows: (1.6000)≈0.200 (1.5875)≈0.133 (1.5750)≈O.067 (1.5625)≈0.003 (1.5562) ≈ 0.029 (1.5500) ≈ 0.060 Therefore, an approximate solution (accurate to 0. OL) of the equation is obtained

Use dichotomy to find a zero point of the function. The reference data are as follows: (1.6000)≈0.200 (1.5875)≈0.133 (1.5750)≈O.067 (1.5625)≈0.003 (1.5562) ≈ 0.029 (1.5500) ≈ 0.060 Therefore, an approximate solution (accurate to 0. OL) of the equation is obtained

(1.5625)≈0.003>0
(1.5562) ≈ 0.029

It is known that the function y = x2 + 2aX + 2 of X is in - 5=

1. Y = x & # 178; + 2x + 2 = (x + 1) & # 178; + 1 when x = 5, y = 37; when x = - 5, y = 17 ﹣ ymax = 37yman = 12, y = x & # 178; + 2aX + 2 = (x + A) & # 178; + 2-A & # 178; when x = 5, y = 10A + 27; when x = - 5, y = - 10A + 27 ﹣ when a ≥ 5, ymax = 10A + 27yman = - 10A + 27; when 0 ≤ a

Given the function y = f (x) = x + 2aX + 2, think ∈ [- 5,5]
When a = - 1, find the maximum and minimum of F (x)

When a = - 1, f (x) = x-2x + 2 = x-2x + 1 + 1 = (x-1) + 1, its axis of symmetry is x = 1, ∵ x ∈ [- 5,5]; when x = 1, f (x) has the minimum value, the minimum value is f (1) = 1; when x = - 5, f (x) has the maximum value, the maximum value is f (- 5) = - 5-1) + 1 = 37

F (x) = 2ax-1 / x, X belongs to (0,1] if the function in (0,1] is an increasing function, find a range

For the derivation of F (x), f '(x) = 2A + 2 / x ^ 3, X belongs to (0,1], f' (x) is ≥ 0 on (0,1]
That is, a ≥ - 1 / x ^ 3, where x belongs to (0,1]
In this interval, - 1 / x ^ 3 is less than or equal to - 1, so a is greater than or equal to - 1
, x ^ (- 3) ≥ 1 ≥ - A, so a ≥ - 1

If the function y = (1 / 2) ^ (x ^ 2-2ax) is an increasing function in the interval (negative infinity, 1), then the range of real number a is

This is a composite function
y=(1/2)^u,u=x²-2ax
Y is a decreasing function of u and an increasing function of X, so u is a decreasing function on (- ∞, 1)
The axis of symmetry of u is x = a with the opening upward
That is, 1 ≤ a
That is, the range of a is [1, + ∞)

Find the maximum and minimum values of the function f (x) = x ^ 2-2x + 1 in the interval [- 2, M]

f(x)=x^2-2x+1=(x-1)^2
Drawing
Adang-2

Find the maximum and minimum values of the function f (x) = x ^ 3-2x ^ 2 + 1 in the interval [- 1,2]

f'(x)=3x^2-4x=x(3x-4)
When x 4 / 3, f '(x) > 0, f (x) increases monotonically
0

The definition field of function f (x) is r, if f (m-n) = f (m) + (n-2m-1) n holds for all real numbers M.N
There is another condition, and f (0) = 0, find f (x)

In the above formula, let m = 0, f (- n) = f (0) + (n-1) n = n ^ 2-n
So f (x) = x ^ 2 + X

Let f (x) = INX + x ^ 2 + ax, if f (x) is an increasing function in its domain of definition, find the value range of A
Please explain in detail

If f (x) = INX + x ^ 2 + ax is defined as (0, + ∞) f '(x) = 1 / x + 2x + AF (x) is an increasing function in its domain, then if x ∈ (0, + ∞), f' (x) ≥ 0 is constant, that is, if 1 / x + 2x + a ≥ 0A ≥ - 2x-1 / X is constant, a ≥ - 2x-1 / x) max ∵ x > 0  2x + 1 / X ≥ 2 √ 2 (if and only if 2x = 1 / x, x = √ 2 / 2, take the equal sign

Given that the definition field of function f (x) is (0, positive infinity) and satisfies 2F (x) + F (1 / x) = (2x-1 / x) INX (1), find the analytic expression and minimum value of F (x),
2) Proof: any x belongs to (0, positive infinity), x + 1 / e ^ X

Solution (1): let x = 1 / x, we get 2F (1 / x) f (x) = (2 / x-x) ln1 / x, we get 4f (x) 2F (1 / x) = 2 (2x-1 / x) LNX from the original formula * 2, we subtract the two formulas to get f (x) = xlnx