Find the maximum or minimum of the following formula, think x & # 178; + 5x + 1 - 2x-6x + 1

Find the maximum or minimum of the following formula, think x & # 178; + 5x + 1 - 2x-6x + 1

When y = x & # 178; + 5x + 1 = (X & # 178; + 5x + 25 / 4) - 21 / 4 = (x + 5 / 2) & # 178; - 21 / 4. X = - 5 / 2, the minimum value is: - 21 / 4, no maximum value. Y = - 2x & # 178; - 6x + 1 = - 2 (x + 3x + 9 / 4) + 13 / 4 = - 2 (x + 3 / 2) & # 178; + 13 / 4

Given that 0 ≤ x ≤ 1, f (x) = x2 − ax + A2 (a > 0), the minimum value of F (x) is m. (1) use a to represent m; (2) find the maximum value of M and the value of a at this time

(1) ∵ f ′ (x) = 2x − a = 2 (x − A2). ① when a > 2, A2 > 1, f ′ (x) ＜ 0, f (x) monotonically decreases on [0, 1], and obtains the minimum value at x = 1, f (1) = 1-A + A2 = 1 − A2. ② when 0 < a < 2, 0 < A2 < 1, Let f ′ (x) = 0, and the solution is x = A2. The list is as follows: from the table, we can see that f (x) obtains the minimum value at x = A2, f (A2) = − A24 + A2, which is also the minimum value. ③ When a = 2, on X ∈ [0, 1], f ′ (x) = 2 (x-1) ≤ 0, the function f (x) monotonically decreases, and the minimum value 0 is obtained at x = 1. In conclusion, M = − A24 + A2, when 0 ＜ a ≤ 2, 1 − A2, when a ＞ 2. (2) ① when 0 ＜ a ≤ 2, m ′ (a) = − 12a + 12 = 1 − A2, when 0 ＜ a < 1, m ′ (a) ＞ 0, the function m (a) monotonically increases; when 1 ＜ a ≤ 2, m ′ (a) ＜ 0, The function m (a) decreases monotonically. When a = 1, m (a) has a maximum value of 14, which is also the maximum value. ② when a > 2, m (a) = 1 − A2 decreases monotonically on (2, + ∞), m (a) < m (2) = 0. In conclusion, only when a = 1, m (a) has a maximum value of 14

Decompose the following formulas 1. - 9 (m-n) &# 178; + (M + n) &# 178; 2.3x & # 178; - one third 3.4 - (X & # 178; - 4x + 2) &# 178;

(m-n) (N-9 (m-n) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x & # 178; - 4x + 2

When x is the value, the formula | x-3 | + | x + 2 | has the minimum value, and the minimum value is obtained

|a|+|b|≥|a+b|
When ab ≥ 0, take the equal sign
So | x-3 | + | x + 2|
=|3-x|+|x+2|≥|3-x+x+2|=5
Take the equal sign, then (3-x) (x + 2) ≥ 0
(x-3)(x+2)≤0
-2≤x≤3
So when - 2 ≤ x ≤ 3, the minimum value is 5

Power series expansion of functions
Let me ask you a question: arctan (1 + x) / (1-x) develops into McLaurin series
Look at the answer, I don't understand why one more quarter Π will be added

arctan((1+x）/(1-x))=arctan(1)+arctan(x)=pi/4+arctan(x)
Then arctan (x) expansion is not a problem

Expansion of function power series
Find the expansion of 1 / (1 + 2x) at x = 0

Because 1 / (1 + x) = 1-x + x ^ 2-x ^ 3 +... + (- 1) ^ (n-1) x ^ n +
So 1 / (1 + 2x) = 1 - (2x) + (2x) ^ 2 - (2x) ^ 3 +... + (- 1) ^ (n-1) (2x) ^ n +
=1-2x+2^2x^2-2^3x^3+...+(-1)^{n-1)2^nx^n+...

Power series expansion of power exponential function

A problem of expanding a function into a power series
The hyperbolic sine SHX = [e ^ x-e ^ (- x)] / 2 is expanded into a power series of X

Then, the even exponential terms which are opposite to each other are eliminated by summation
x + x^3/3!+ x^5/5!+ x^7/7!+ x^9/9!+ x^11/11!+ x^13/13!+ x^15/15!+...+x^(2n-1)/(2n-1)!+...

The power series of F (x) = 1 / (1 + x) at x0 = 0 is 1 / (x + 1) =?

Let f (T) = 1 / (1-T) be f (T) = 1 + T + T ^ 2 + T ^ 3 + at 0 +t^n+…… = ∑t^n (n=0,1,2,3.)
Let t = - x, then 1 / (1-T) = 1 / (1 + x) = ∑ T ^ n = ∑ (- x) ^ n (n = 0,1,2,3.)
That is, f (x) = 1 / (1 + x) is expanded at 0 as f (x) = ∑ (- x) ^ n (n = 0,1,2,3.)

The function f (x) = cosx is expanded into a power series of (x-1), and the interval of the expansion is obtained,

Let t = X-1
Then x = t + 1
cosx=cos(t+1)=costsin1-sintcos1
=sin1[1-t^2/2!+t^4/4!-...]-cos1[t-t^3/3!+t^5/5!-..]
=sin1-(cos1)t-(sin1)t^2/2!+(cos1)t^3/3!-.
This is the power series about X-1. The convergence domain is r