Find the range of function y = 3x & # 178; + (1 / 2x & # 178;)

Find the range of function y = 3x & # 178; + (1 / 2x & # 178;)

y≥2√(3x^2*[1/(2x^2)]
=2√(3/2)
=√6
The range is [√ 6, + ∞)

Given the function f (x) = 2x ^ 2-4x + 1 and X belongs to [- 2,6], find the maximum value of F (x) and the range of F (x) = x ^ 2-4x-2 (x belongs to R)

【1】 The axis of symmetry of F (x) = 2x ^ 2-4ax + 1 is x = a
Discussion:
(1) When a ≤ - 2, the function increases at [- 2,6]
So when x = - 2, the function has a minimum value of 9 + 8A
When x = 6, the function has a maximum value of 73-24a
(2) When a ∈ [- 2,6],
When x = a, the minimum value is - 2A & sup2; + 1
For the maximum value, we need to compare which side is close to the axis of symmetry, that is, which side is close to the axis of symmetry 2 of [- 2,6]
So when - 2 ≤ a ≤ 2, the axis of symmetry x = a is closer to - 2
When x = 6, the maximum value is 73-24a
When 2 ≤ a ≤ 6,
When x = - 2, the maximum value is 9 + 8A
(3) When a ≥ 6, the function decreases at [- 2,6]
When x = - 2, the maximum value is 9 + 8A
When x = 6, there is a minimum value of 73-24a
【2】 For the function f (x) = 2x ^ 2-4ax + 1 = 2 (x-a) & sup2; - 2A & sup2; + 1,
When x belongs to R, the minimum value of the function at the symmetry axis is - 2A & sup2; + 1,
So the range of F (x) is [- 2A & sup2; + 1, + ∞)

An abstract function problem in senior one
It is known that even function f (x) has f (x + 3) = - 1 / F (x) for any x ∈ R, and if x ∈ (- 3, - 2), f (x) = 2x, the value of F (113.5) can be obtained
I don't know which step. Unexpectedly, it's stuck
F (x + 3 + 3) = - 1 / F (x + 3) = - 1 / [- 1 / F (x)] = f (x), that is, f (x) = f (x + 6),
The period T = 6,
There's no problem in the back. The answer is 1 / 5

If x ∈ (2,3), the analytic expression of the function is f (x), then - x ∈ (- 3, - 2), indicating that f (- x) = - 2x. Because the function is even, f (x) = f (- x) = - 2x. According to f (x + 3) = - 1 / F (x), the transformation is: F (x + 3) * f (x) = - 1, indicating that the independent variable + 3, the value of the function becomes negative reciprocal