Find the monotone interval, concave convex interval, extremum, inflection point and asymptote of function y = x + X / (x ^ 2-1)

Find the monotone interval, concave convex interval, extremum, inflection point and asymptote of function y = x + X / (x ^ 2-1)

y=x^3/(x^2-1)
y'=[3x^2(x^2-1)-2x^4]/(x^2-1)^2=x^2(x^2-3)/(x^2-1)^2
From y '= 0, we get: x = 0, √ 3, √ 3, where when x = 0, the neighborhood around y' is invariant, that is, x = 0 is not an extreme point
Monotonically increasing interval: x > √ 3, or X

As y = 1 + x ^ 2 of x ^ 2 concave convex interval, inflection point, increase and decrease interval, extreme value

Y ′ = 2 * x / (1 + X? 2) y ″ = 2 * (1-3 * x? 2) / (1 + X? 2) 3 y ′ = 0, x = 0, y ′≥ 0, y increases monotonically on [0, + ∞), y ″ = 0, x ″ = 0, x = ± √ (1 / 3), and = ± √ (1 / 3) is y

Find the monotone interval, extremum, inflection point and concave convex interval of Ln (1 + x ^ 2)

Let f (x) = ln (1 + x ^ 2)
Then f '(x) = 2x / (1 + x ^ 2), f' '(x) = 2 (1-x ^ 2) / (1 + x ^ 2) ^ 2
When x > 0, f '(x) > 0
When x

The known function f (x) = LNX ax + 1 − a X-1 (a ∈ R), when a ≤ 1 The monotonicity of F (x) is discussed

f′(x)＝1
x−a−1−a
x2=-ax2−x+1−a
x2=-[ax+(a−1)](x−1)
x2（x＞0），
Let g (x) = ax2-x + 1-A,
① When a = 0, G (x) = - x + 1; when x ∈ (0, 1), G (x) > 0, f ′ (x) < 0, the function f (x) decreases monotonically;
When x ∈ (1, + ∞), G (x) < 0, f ′ (x) > 0, the function f (x) increases monotonically;
② When < 0 a
When f ′ (x) = 0, X1 = 1, X2 = 1
A-1
a-1＞1＞0，
The list is as follows:
It can be seen from the table that the function f (x) is in the interval (0, 1) and (1)
A − 1, + ∞) decreases monotonically on the interval (1,1)
A − 1) increases monotonically;
③ When a = 1
When f ′ (x) ＜ 0, the function f (x) decreases monotonically at (0, + ∞);
④ When a < 0, due to 1
If A-1 < 0, the function f (x) decreases monotonically on (0, 1) and increases monotonically on (1, + ∞)
To sum up: when a ≤ 0, the function f (x) decreases monotonically on (0, 1) and increases monotonically on (1, + ∞)
When a = 1
When 2, the function f (x) decreases monotonically on (0, + ∞)
When 0 < a < 1
When 2, the function f (x) is in the interval (0, 1) and (1)
A − 1, + ∞) decreases monotonically on the interval (1,1)
A − 1)

The known function f (x) = LNX ax + (1-A) / X (0)

If f (x) = LNX ax + (1-A) / XF '(x) = 1 / x-a + (A-1) / x? = [- ax? + X + (A-1)] / x? = - a [x? - 1 / A * x + (1 / A-1)] / x? = - A (x-1) [x - (1 / A-1)] / x? When a = 1 / 2, 1 / A-1 = 1F' (x) = - 1 / 2 (x-1) / x? ≤ 0 is constant

Let f (x) = 1 / 2x ^ 2-ax + (A-1) LNX, a > 1 Proof: if A-1

prove:
Consider the function g (x) = f (x) + x = 1 / 2x? - ax + (A-1) LNX + X
Then G '(x) = x - (A-1) + [(A-1) / x] ≥ 2 √ [x · (A-1) / x] - (A-1) = 1 - [√ (A-1) - 1] 2
Due to 1

The function f (x) = AX2 + 2x + 1, G (x) = LNX (1) Let f (x) = f (x) - G (x), find the necessary and sufficient conditions of two extremum points of F (x) (2) It is proved that when a ≥ 0, the inequality f (x) ≥ g (x) always holds

(1) The function f (x) = AX2 + 2x + 1, G (x) = LNX,
∴F（x）=f（x）-g（x）=ax2+2x+1-lnx，
Its definition domain is (0, + ∞)
∴F‘(x)＝2ax+2−1
x=2ax2+2x−1
x，
There are two extreme points of F (x),
The equation 2ax2 + 2x-1 = 0 has two unequal positive roots,
Qi
△＝4+8a＞0
x1+x2＝−1
a＞0
x1•x2＝−1
2a＞0 ，
The solution is − 1
2＜a＜0，
The necessary and sufficient condition for f (x) to have two extreme points is − 1
2＜a＜0．
(2) It is proved that the inequality f (x) ≥ g (x) is always true if and only if:
F (x) = AX2 + 2x + 1-lnx ≥ 0 holds on (0, + ∞),
That is, a ≥ LNX − (2x + 1)
X2 always holds on (0, + ∞)
Let H (x) = LNX - (2x + 1), then h ′ (x) = 1
x−2＝1−2x
x，
When x ∈ (0, 1
2) When h ′ (x) > 0,
When x ∈ (1)
2, + ∞), H ′ (x) ＜ 0
∴x＝1
At 2, H (x) max = ln1
2−2＜0，
So x ∈ (0, + ∞) has LNX − (2x + 1)
x2＜0，
When a ≥ 0, a ≥ LNX − (2x + 1)
X2 always holds on (0, + ∞),
That is, when a ≥ 0, the inequality f (x) ≥ g (x) holds

On the function f (x) = logax + 1 The monotonicity of X-1 (a > 0 and a ≠ 1) on (1, + ∞) is proved

Let u2-u1 = x2 + 1x2-1-x1 + 1x1-1 = (x2 + 1) (x1-1) (x1-1) (x1-1) (x1 + 1) (x1 + 1) (x1 + 1) (x1-1) (x2-1) (x2-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1) (x1-1), then u2-u1 = x2 + 1x2 + 1x1-1 + 1x1-1 + 1x1-1 + 1x1-1 = (x2) (x1 + 1) (x1 + 1) (x1 + 1) (x1-1) (x1-1) (x1-1) ((x2-x2) (x2-1) (x1-1) < 0

The known function f (x) = loga [(4 + x) / (4-x)] + 1 / X (0

On the same continuous interval of the domain (- 4,0) ∪ (0,4), there is m

Given the function f (x) = LNX + 2x-6, (1) it is proved that f (x) is an increasing function in its domain of definition, (2) It is proved that f (x) has and has only one zero point,

1. The domain of function definition is x > 0
Y '= 1 / x + 2 > 0
Y '' = - 1 / x ^ 2 < 0. The function is convex
2.f'(x)=1/x+2>0,
So f (x) increases monotonically,
Because x tends to 0 - ∞,
When x = e, f (x) > 0,
So f (x) has only one zero point