If the derivative of the power function y = f (x) is known to pass through the point (1,1 / 2), then f (2)=

If the derivative of the power function y = f (x) is known to pass through the point (1,1 / 2), then f (2)=

Analysis:
Let the power function be y = x Λ α,
Then y '= α * x Λ (α - 1)
Because the derivative is over (1,1 / 2), y 'is taken in
1/2＝α*1＝α
Therefore, the power function is
y＝x∧1/2
∴f(2)＝2∧1/2＝√2

Ln (x power of E + 1)?
A little ambiguity is ln [(x power of E) + 1]

Let u = e ^ x + 1, then y = ln U
Y'x = y'u times u'x
=(LN U) 'times (e ^ x + 1)'
=1 / u times e ^ x
Then we substitute u = e ^ x + 1 to get
y'=e^x/(e^x+1)

Through the formula, the quadratic function is changed into y = a (X-H) ^ 2 + K1. Y = x ^ 2-4x + 22. Y = - 2 / 2 x ^ 2 + 2x + 1

y=x^2-4x+2=(x-2)^2 -2
y=-(x^2)/2+2x+1=(-1/2)[x^2-4x]+1=(-1/2)[(x-2)^2-4]+1=(-1/2)(x-2)^2+3

The quadratic trinomial formula x2-4x + 1 was formulated to get ()
A. （x-2）2+3B. （x-2）2-3C. （x+2）2+3D. （x+2）2-3

∵ x2-4x + 1 = x2-4x + 4-4 + 1, x2-4x + 1 = (X-2) 2-3, so B

The formula is x square - 4x + 4 = 1 + 4 (x -) square = 5 (x -) = ± √ 5 x = ± √ 5
By solving the equation x square - 4x - 1 = 0 with the method of matching, the formula is x square - 4x + 4 = 1 + 4

X squared - 4x + 4 = 1 + 4
(X-2) square = 5
（x-2）=±√5
X=2±√5

The quadratic trinomial formula x2-4x + 1 was formulated to get ()
A. （x-2）2+3B. （x-2）2-3C. （x+2）2+3D. （x+2）2-3

∵ x2-4x + 1 = x2-4x + 4-4 + 1, x2-4x + 1 = (X-2) 2-3, so B

Compare the size of 3x2 - 2x + 7 and 4x2 - 2x + 7, please write the process

∵（4x2-2x+7）-（3x2-2x+7）=x2≥0，∴4x2-2x+7≥3x2-2x+7．

Compare the size of 3x ^ 2 + 2x + 5 and x ^ 2-4x-2

3x^2+2x+5-(x^2-4x-2)=2x^2+6x+7=2(x+1.5)^2+2.5>=2.5
So 3x ^ 2 + 2x + 5 is greater than x ^ 2-4x-2

X ^ 2-4x + 1 is transformed into (x + 5) ^ 2 + K (where K and H are constants)

x²-4x+1
=x²-4x+4-4+1
=（x-2）²-3

What is the form of y = - 2x & sup2; - 4x + 5 to y = a (X-H) & sup2; + k?

y = -2x² - 4x + 5
= -2x² - 4x - 2 + 7
= -2(x + 1)² + 7