Compare the size of tan2 and tan9

Compare the size of tan2 and tan9

1=57°

Given a = Log1 / 52, B = (1 / 2) ^ 0.6, C = 5 ^ (1 / 2), then the size relationship of a, B, C is?
RT

a=log1/5

Help compare log 1 / 4 bottom 8 / 7 and log 1 / 5 bottom 6 / 5 size

The first formula is 7 / 8 divided by 1 / 4, which is 7 / 2 of LN. The second formula can be changed to 25 / 6 of LN. Obviously, the second one is larger, and LN is an increasing function

The size of Log1 / 4 bottom 3 / 4 is compared with log5 bottom 6

The size of Log1 / 4 bottom 3 / 4 is compared with log5 bottom 6
Log1 / 4 bottom 3 / 41
So Log1 / 4 bottom 3 / 4 < log5 bottom 6

log4(3x＋1）＋log4(x－1)＜1

log4[(3x+1)(x-1)]＜log4 4
(3x+1)(x-1)＜4
So - 1 < x < 5 / 3

Given the function f (x) = 3x-x2, find the number of real roots of the equation f (x) = 0 in the interval [- 1,0]

∵ f (- 1) = 3-1 - (- 1) 2 = - 23 ＜ 0, f (0) = 30-02 = 1 ＞ 0, ∵ f (- 1) · f (0) ＜ 0. The graph of function f (x) on [- 1,0] is continuous, the equation f (x) = 0 has real roots in [- 1,0]. The function f (x) = 3x-x2 is an increasing function on [- 1,0], and the equation f (x) = 0 has only one real root on [- 1,0]

Given the function f (x) = 3x-x2, find the number of real roots of the equation f (x) = 0 in the interval [- 1,0]

∵ f (- 1) = 3-1 - (- 1) 2 = - 23 ＜ 0, f (0) = 30-02 = 1 ＞ 0, ∵ f (- 1) · f (0) ＜ 0. The graph of function f (x) on [- 1,0] is continuous, the equation f (x) = 0 has real roots in [- 1,0]. The function f (x) = 3x-x2 is an increasing function on [- 1,0], and the equation f (x) = 0 has only one real root on [- 1,0]

If x = 3 is a root of the equation f (x) = 0, then the minimum number of roots of F (x) = 0 on the interval (0,10) is?
Such as the title

They are 2,4,8 respectively,

The number of roots of the equation f (x) = x & sup2; + A / x = x is discussed

Equivalent to seeking
The root of x ^ 3-x ^ 2 + a = 0
First, because x is the denominator, f (x) is undefined when x = 0
Then x ^ 3-x ^ 2 is a monotone function on (negative infinity, 0), with values from (negative infinity to 0)
It is a monotone decreasing function on (0,2 / 3) with values from (0, - 4 / 27)
Where (2 / 3, positive infinity) is a monotone increasing function, the value is from (- 4 / 27, positive infinity)
So, the conclusion is
When
a

Given the function f (x) = x2 + ax + 3, if f (x) ≥ A is [- 2,1] constant for X, find the value range of real number a

a≤x²+ax+3
a(1-x)≤x²+3,
When x = 1, the above formula is:
0 ≤ X & # 178; + 3, = = < A is all real numbers;
When X0
a≤(x²+3)/(1-x)
Let 1-x = t
X = 1-T, X & # 178; = T & # 178; - 2T + 1 and - 2 ≤ 1-t0t = 2
G '(T) is a left negative right positive corresponding function near T = 2, and G (T) is a left minus right increasing function, so the minimum value of the function is:
g(min)=g(2)=2
If a ≤ 2 and a ∈ R, then the intersection is obtained
a≤2