The function f (x) = 1 / 2x ^ 2-alnx is known. When a = - 1, it is proved that the function f (x) increases monotonically on (0, + infinity). 2 find the extremum of F (x)

The function f (x) = 1 / 2x ^ 2-alnx is known. When a = - 1, it is proved that the function f (x) increases monotonically on (0, + infinity). 2 find the extremum of F (x)

1）a=-1,f(x)=1/2* x^2+lnx
The domain is x > 0
In the domain, 1 / 2 x ^ 2 and LNX are monotone increasing functions, so f (x) is monotone increasing in the domain
2) Because f (x) increases monotonically, there is no extremum

It is known that f (x) = x + A / x ^ 2 + BX + 1 is an odd function (1) to find the value of a and B, and (2) to find the monotone interval of F (x)

Reserve knowledge:
1) Odd function:
Let the domain of definition of the function y = f (x) be D, and d be a symmetric set of numbers about the origin. If for any X in D, there is x ∈ D, and f (- x) = - f (x), then this function is called an odd function
2) Derivative:
In general, let y = f (x) be defined near the point x0 (x0-a, x0 + a);
When the increment of the independent variable Δ x = x-x0, Δ x → 0, the limit of the ratio of the function increment Δ y = f (x) - f (x0) to the increment of the independent variable exists and is limited, that is to say, the function f is differentiable at x0, which is called the (or rate of change) of F at x0
If the function f is differentiable at every point in the interval I, a new function with I as the domain of definition is obtained, denoted as f (x) 'or y', which is called the derivative of F for short
Geometric meaning of derivative f '(x0) of function y = f (x) at x0: it indicates the tangent slope of function curve at P0 [x0, f (x0)] (geometric meaning of derivative is the tangent slope of function curve at this point)
In general, we get a rule to judge the monotonicity of a function by its derivative: let y = f (x) be differentiable in (a, b). If f '(x) > 0 in (a, b), then f (x) increases monotonically in this interval (the tangent slope of the point increases, and the function curve becomes "steep" and ascending)

It is known that f (x) = (x-a) / [x & sup2 + BX + 1] is an odd function (1) to find the value of a and B and (2) to find the monotone interval of F (x)

(1) Odd function, f (x) = - f (- x), that is: (x-a) / [x ^ 2 + BX + 1] = - (- x-a) / [x ^ 2-bx + 1] cross multiplication of numerator and denominator, x ^ 3-bx ^ 2 + x-ax ^ 2 + abx-a = x ^ 3 + BX ^ 2 + X + ax ^ 2 + ABX + A, a = b = 0 (2) f (x) = x / (x ^ 2 + 1), derivative = (1-x ^ 2) / (x ^ 2 + 1) ^ 2, satisfies the above formula greater than

Given that f (x) = (x ^ 2 + 1) / (BX + C) is an odd function and f (1) = 2, find the value of B and C. find the monotone interval of F (x)

Let me make the following points in detail: ∵ f (x) = (x ^ 2 + 1 / (BX + C) is an odd function ∵ there is f (- x) = - f (x) - -------- ① ∵ from ①, we can establish the equation: (- x) ^ 2 + 1 / (b (- x) + C) = - (x ^ 2 + 1 / (BX + C)) - C = C, ∵ C = 0 ∵ f (x) = (x ^ 2 + 1 / BX) and ∵ f (1) = 2 ∵ 2 / b = 1 ∵ f (x) = (x ^ 2 / x) = x + 1 / x, let x1, (x2) and (x1 ＜ x2 {X2 (x2 2x \\\\\\\\\\\\\\\\\\\\\\ (x2) - f (x (x2) - f (x (x (x2) - 2-x1-1-1-1-1-1-1-1-1-1-1-1 / 1 / x1-1 / x1-1 / X1 (x1-x \ (x2 \\\\\\\\\\\\\\\ifx belongs to [- 1,0) or (0,1], then, The monotone interval of F (x) is decreasing. When x belongs to (- ∞, - 1) or (1, ∞), x1x2-1 ＞ 0 ∵ x2-x1 ＞ 0, x1x2 ＞ 0, x1x2-1 ＞ 0 ∵ x2-x1 (x1x2-1 / x1x2) ＞ 0, that is, f (x2) - f (x1) ＞ 0, f (x2) ＞ f (x1) ≠ when x belongs to (- ∞, - 1) or (1, ∞), the monotone interval of F (x) is increasing. If you don't understand, you can ask again

Given the equation x ^ lgx = x ^ 3 / 100, then the value of X is?
Such as the title
Ask for detailed explanation

Lgx 2; - LG (x 3) = LG (x 2 / x 3) = loga a = (x 2 / x 3) for (x) when x = 4 16 / 7, the value range of X is 3 / 2 < a < 16 / 7

How to divide the 4.8m long and 3M wide rectangular carpet into two congruent figures and lay them in a 4m and 3.6m long and wide room?

It is known that ABC is the trilateral length of △ ABC, and a & sup2; + B & sup2; + C & sup2; + 50 = 6A + 8b + 10C. It is proved that △ ABC is a right triangle

A & sup2; + B & sup2; + C & sup2; + 50 = 6A + 8b + 10C can be changed into (A-3) & sup2; + (B-4) & sup2; + (C-5) & sup2; = 0
Then a = 3, B = 4, C = 5

The area of a triangle is 12.6 square centimeter, the bottom is 8.4 centimeter, seek height

12.6×2÷8.4
=3 cm

Given the point m (- 2,4) and the parabola with focus f y = 1,8x2, find a point P on this parabola to minimize the value of | PM | + | PF |

Define by parabola: | PF | = | PP '| desire | PM | + | PF |, P, p', m should be three points collinear, then the abscissa of P point is - 2 happy New Year! Ask: sorry, I didn't understand your answer. Why not use M.P.F three points collinear? And my answer is different from yours Answer: the minimum value of PM + pf first considers point M

If the value of algebraic formula 8-x4 is not less than that of algebraic formula 3x + 5, then the value range of X is______ ．

∵ 8-x4 ≥ 3x + 5, ∵ shift term combined with similar term, - 134x ≥ - 3, coefficient 1, X ≤ 1213