It is proved by limit definition that function 1 / (1 + 2x) is equal to 1 / 3 when x tends to 1

It is proved by limit definition that function 1 / (1 + 2x) is equal to 1 / 3 when x tends to 1

consider
| 1/(1+2x) - 1/3 |
=| (3-1-2x)/3(1+2x) |
=| (2-2x)/3(1+2x) |
=(2/3)* | x-1 |/| 1+2x |
0, when | X-1|

The function is defined at 0, f (0) = 1, f '(0) = 1, limf (x) / x =? (x tends to 0)
Is it = 1

No, because if you don't input a good formula, you won't be rewarded for the process. Starting from the definition of derivative, write the expression of F '(0) = 1. If you combine f (0) = 1, you will get that the obtained limf (x) / X is nonexistent or infinite

Let f (x) be second-order differentiable at x = A and limf '(x) / (x-a) = - 1, then () a.x = a is f (x)
Let f (x) be second-order differentiable at x = A and limf '(x) / (x-a) = - 1, then ()
A. X = a is the minimum of F (x)
B. X = a is the maximum point of F (x)
C. (a, f (a)) is the inflection point of the curve y = f (x)
D. X = a is not the extreme point of F (x), and a, f (a) is not the inflection point of the curve y = f (x)

From Lim f '(x) / (x-a) = - 1, f' (a) = 0, and F "(a) = - 1

Find the tangent equation and normal plane of curve X ^ 2 + y ^ 2 + Z ^ 2 = 6, Z ^ 2 + y ^ 2-x ^ 2 = 4 at point (1,1,2)

Believe in yourself, you can do it!

Find out the curve X = t, y = t, z = T3, so that the tangent at this point is parallel to the plane x + 2Y + Z = 4

The tangent slope (derivation) of the curve X = t, y = T ^ 2, z = T ^ 3 is x = 1, y = 2T, z = 3T ^ 2. The tangent is parallel to the plane x + 2Y + Z = 4. The product of the tangent slope and the normal vector of the plane is 0 1 * 1 + 2T * 2 + 3T ^ 2 * 1 = 0 t = - 1 or - 1 / 3, which is substituted into the linear equation x = - 1, y = 1, z = - 1, or x = - 1 / 3, y = 1 / 9, z = - 1 / 2

The line 3x-2y + 4 = 0 and 6x-4y-5 = 0 are two parallel tangents of a circle. What is the area of the circle

The diameter of a circle is the distance between two straight lines
Take a point on 3x-2y + 4 = 0
For example (0,2)
Then the distance from him to another straight line is the diameter
So d = | 0-8-5 | / √ (6 & sup2; + 4 & sup2;) = √ 13
So area = 13 π / 4

If the line l1:3x-y + 4 = 0 and the line l2:6x-2y-1 = 0 are two tangents of a circle, then the area of the circle --?

The throne, virtue and rule are infinite
Count me as an almond
Is it you, hopeless?
It's like staring at something strange
——You have salt in your eyes, I said
So, ha ha

The parabola C: x ^ 2 = 4Y passing through point (4,2) is tangent L1 and L2 of parabola at two points a and B respectively
When s △ ABC = 28 root sign 7, calculate the coordinates of N point
It's a triangle ABN

Let the equation of the straight line L passing through the point P (4,2) be y = K (x-4) + 2 (it is obviously impossible to be the equation x = 4, because there is only one intersection point between the straight line x = 4 and the parabola C: y = 1 / 4 * x ^ 2), which is combined with the parabola equation y = 1 / 4 * x ^ 2
1 / 4 * x ^ 2 = K (x-4) + 2, i.e
x^2-4kx+8(2k-1)=0.
Let a (x1,1 / 4 * X1 ^ 2) and B (x, 2,1 / 4 * X1 ^ 2) have
x1+x2=4k ①
x1*x2=8(2k-1) ②
If the derivative of the function y = 1 / 4 * x ^ 2 is y '= x / 2, then the slope of the tangent passing through a point (x, y) on the parabola C is y' = x / 2
The linear L1 equation is Y-1 / 4 * X1 ^ 2 = X1 / 2 * (x-x1) ③
The equation of line L2 is Y-1 / 4 * x2 ^ 2 = x2 / 2 * (x-x1) ④
③ - 4
-1/4*(x1^2-x2^2)=x/2*(x1-x2)-1/2*(x1^2-x2^2)
Because x1 ≠ X2, so x1-x2 ≠ 0, then the above formula can be reduced to
x=(x1+x2)/2
So y = 1 / 4 * X1 ^ 2 + X1 / 2 * (x-x1) = X1 * x2 / 4
Combining with ① and ②, it is easy to get that the coordinates of intersection n of line L1 and line L2 are n (2k, 4k-2)
The equation of line L passing through point P (4,2) is kx-y + 2-4k = 0, then the distance from point n to line L is 0
s=[k*2k-(4k-2)+2-4k]/√(k^2+1)=(2k^2-8k+4)/√(k^2+1)
AB=√[(x2-x1)^2+(1/4*x2^2-1/4*x1^2)^2]=√{[(x1+x2)^2-4x1x2][1+1/16*(x1+x2)^2]}
=4√[(k^2+1)(k^2-4k+2)]
So from s △ ABN = 28 √ 7
1/2*4√[(k^2+1)(k^2-4k+2)]*(2k^2-8k+4)/√(k^2+1)=28√7
It is reduced to (k ^ 2-4k + 2) ^ (3 / 2) = 7 ^ (3 / 2)
So K ^ 2-4k + 2 = 7
The solution is k = - 1 or K = 5
Then the coordinates of N point are (- 2, - 6) or (10,18)

A. B is the moving chord passing through the focus of the parabola x2 = 4Y, and the straight lines L1 and L2 are two tangent lines of the parabola tangent to a and B respectively, then the ordinate of the intersection point of L1 and L2 is ()
A. -1B. -4C. −14D. −116

Take a special case when ab ⊥ Y axis, then a (- 2,1), B (2,1), the tangent equation passing through point a is Y-1 = - (x + 2), that is, x + y + 1 = 0. Similarly, if the tangent equation passing through point B is x-y-1 = 0, then solve the equations x + y + 1 = 0x − y − 1 = 0, and the intersection point of L1 and L2 is (0, - 1)

x. Y ∈ R, try to compare the size of x ^ 2 + 5Y ^ 2 + 1 and 2Y (2x + 1)

Methods: do the difference method
Original formula = x ^ 2 + 5Y ^ 2 + 1-4xy-2y
=4y^2-4xy+x^2+y^2-2y+1
=(2y-x)^2+(y-1)^2
therefore
x^2+5y^2+1>=2y(2x+1)