What is the minimum value of formula | 2x + 1 | + 1, where x =?

What is the minimum value of formula | 2x + 1 | + 1, where x =?

The absolute value is greater than or equal to 0
When 2x + 1 = 0, that is, when x = - 1 / 2
There is a minimum value of 1

When the absolute value of formula 2x-1 = 2, taking the minimum value, X=

That is, 2x-1 = plus or minus 2
When 2x-1 = 2
2x=3,
x=1.5
When 2x-1 = - 2
2x=-1
x=-0.5
So x = - 0.5

Is the second power derivative of E 0

E ^ 2 is a constant and its derivative is 0

In this paper, f (x) = (x-1) / (4-x) is expanded into a power series of X-1, and the n-th derivative f ^ (n) (1) of F (x) at x = 1 is obtained
The first question has been asked. The second question is how to ask?
(x-1) / (4-x) = 1 / 3 (x-1) + (x-1) ^ 2 / 3 ^ 2 + (x-1) ^ 3 / 3 ^ 3 +... + (x-1) ^ n / 3 ^ n +
The answer is f ^ (n) (x) = n! / 3 ^ n + (n + 1) n... 2 (x-1) / 3 ^ (n + 1) + (n + 2) n... 3 (x-1) ^ 2 / 3 ^ (n + 2) +
How did you get it? Please be more specific

Now that the power series is known, it is easy to find the n-th derivative of the power series
N-order derivative of (x-1) ^ n = n! (x-1) ^ (n + k) = (n + k) (n + k-1)... (K + 1) x ^ k
f^(n)(x)=n!/3^n+(n+1)n...2(x-1)/3^(n+1)+(n+2)n...3(x-1)^2/3^(n+2)+...
f^(n)(1)=n!/3^n

In order to find some expansions of power series, do you want to write out the steps of solving Maclaurin series? Or do you want to remember some commonly used expansions of power series? Maybe I don't express it clearly, but in the end, I want to solve the problem of expansion of power series I hope the great God will teach me!

These are all necessary, some problems are simple ones, and the steps of solving problems must be remembered, while the common ones are for solving problems faster and can be remembered as much as possible

Solving f (n) (1) by using the power series expansion of F (x) = 1 / x ^ 4 + 4x + 3 in x0 = 1

f（x）=1/x^4+4x+3=1/(x+3)(x+1)=1/2*[1/(x+1)-1/(x+3)]=1/2*[1/[2+(x-1)]-1/[4+(x-1)]]=1/4*1/[1+(x-1)/2]-1/8*1/[1+(x-1)/4]=1/4Σ(-1)^n*((x-1)/2)^n-1/8Σ(-1)^n*((x-1)/4)^n=1/2Σ(-1)^n*[1/2^(n+1)-1/4^(n+1)]...

Let y = x & # 178; - 4x + 1 be written as y = a (X-H) &# 178; + K,

y=x²-4x+1
y=x²-4x+4-3
y=(x-2)²-3

The quadratic function y = - 14x2-x + 3 is formulated as y = a (X-H) 2 + K______ The vertex coordinates of the quadratic function image are______ ．

Y = - 14x2-x + 3 = - 14 (x2 + 4x) + 3 = - 14 (x + 2) 2 + 4, that is, y = - 14 (x + 2) 2 + 4, the vertex (- 2, 4). So the answer is: y = - 14 (x + 2) 2 + 4, (- 2, 4)

How to change the quadratic function y = 1 / 4x ^ 2 + X-1 into y = a (X-H) ^ 2 + k?

y=1/4(x^2+4x)-1
=1/4(x^2+4x+4)-1-1
=1/4(x+2)^2-2

1. The quadratic function y = x ^ 2-4x + 3 is transformed into the form of y = a (X-H) ^ 2 + K
2. If the vertex of the parabola y = 2x ^ 2 + BX + 8 is on the X axis, then B = ()
3. The coordinate of the intersection of the straight line y = 2x + 2 and the parabola y = x ^ 2 + 3x is ()
4. If the image of quadratic function y = (M + 8) x ^ 2 + 2x + m ^ 2-64 passes through the origin, then M = ()
5. Quadratic function y = x ^ 2-4x-3, if - 1 ≤ x ≤ 6, then the value range of Y is ()
(it's better to have a process)

1,y=（x-2）^2-1
2, B = 4 or - 4
3,(1,4),(-2,-2)
4, M = 8 or - 8
5, the range of Y (- 7,9)