If f (x) = x & # 178; + 2x * f '(1), then f' (0) is equal to?

If f (x) = x & # 178; + 2x * f '(1), then f' (0) is equal to?

F '(1) is a constant
So f '(x) = 2x + 2F' (1)
Let x = 1
f'(1)=2+2f'(1)
f'(1)=-2
So f '(0) = 0 × 2 + 2 × (- 2) = - 4

Given f (2x + 1) = x & # 178; - 2x, then f (3 =)

Solution 1: let 2x + 1 = t, then x = (t-1) / 2F (T) = (t-1) &# / 4 - (t-1) f (3) = 1-2 = - 1 solution 2: F (2x + 1) = x2-2x, let y = 2x + 1x = (Y-1) / 2F (y) = [(Y-1) / 2] &# / 178; - (Y-1) = y & # / 178 / 4-3y / 2 + 5 / 4, so f (x) = x & # / 178 / 4 + 3x / 2 + 5 / 4f (3) = f (2 * 1 + 1) = 1 & # - 178; - 2 * 1

Let A1 > 0, an + 1 = 3 (1 + an) / 3 + an (n = 1,2..) find the limit of an when n tends to infinity. How to see 0 in the process of finding the limit

From A1 > 0, we can see that A2 > 0, so A3 > 0, A4 > 0. That is to say, for all a (n), there is a (n) > 0, so for all n = 1,2,. A (n + 1) = 3 * (1 + A (n)) / (3 + a (n))

Simplify 3x ^ 2 * (- 2XY) ^ 2-x (x ^ 2Y ^ 2-2x) 2A * (a ^ 2 + 3a-2) - 3 (a ^ 3 + 2A ^ 2-A + 1)

The original formula = 3x * 4x4x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\#179; - 6A & #178; + 3a-3 = - A-3

The two zeros of function f (x) = xsquare - (K-2) x + ksquare + 3K + 5 are the value range of X1 X2 to find t = X1 square + x2 square

△=(k-2)²-4(k²+3k+5)=-3k²-16k-16>0
D-4

In the image of quadratic function y = x2-6x + 6, when y decreases with the increase of X, the value range of X is

y=x2-6x+6
=(x-3)²-3
The axis of symmetry of a parabola is a straight line x = 3
Since the opening of the parabola is upward, then on the left side of the symmetry axis, y decreases with the increase of X,
Therefore, when y decreases with the increase of X, the value range of X is x < 3

Is it wrong to write decimal before root sign? For example, 0.5 root sign 5
Let's say whether it's wrong

It's OK
There is no miscalculation
0.5 root 2 is root 2 / 2
If you write like this, the general teacher will ask you to write root 2 / 2

To solve the equations: 5x + 6y = 162x − 3Y = 1

5x + 6y = 16 & nbsp; & nbsp; ① 2x − 3Y = 1 & nbsp; & nbsp; ②, ① + ② × 2, 9x = 18, x = 2, 4-3y = 1, y = 1, so the solution of the equations is x = 2Y = 1

4 / 5x-1 / 5 = 14 / 5

4x-1 = 14 (remove the denominator and multiply by 5 on both sides of the equation)
4X = 14 + 1 (move to, change sign)
4X = 15 (merge congeners)
X = 3.75 or 15 / 4

The line L passing through the point m (- 2,0) and the ellipse x ^ 2 + 2Y ^ 2 = 2 intersect at two points P1 and P2. The midpoint of the line p1p2 is p. suppose the slope of the line L is K1 and the slope of the line OP is K2, then the value of K1 * K2 is

Let P1 (x1, Y1), P2 (X2, Y2), P (x0, Y0)
Then X1 ^ 2 + 2y1 ^ 2 = 2
x2^2+2y2^2=2
Then (x1 ^ 2-x2 ^ 2) + 2 (2y1 ^ 2-2y2 ^ 2) = 0
（x1+x2）*（x1-x2）+2（y1+y2）*（y1-y2）=0 ①
And 2x0 = X1 + X2, 2y0 = Y1 + Y2
k1=(y1-y2)/(x1-x2)
That is, y1-y2 = K1 (x1-x2)
k2=(y0-0)/(x0-0)
That is, Y0 = k2x0
In this way, we can substitute ①
2x0(x1-x2)+2*2k2x0*k1(x1-x2)=0
1+2k1k2=0
k1k2=-1/2