The tangent equation at the point (1,2) of the cubic + 1 Curve y = x is The key is how to find the derivative, how to find y 'what's the formula of derivative ~ ~ urgent

On the curve, so it's the tangent point y '= 3x squared, so x = 1, the tangent slope k = y' = 3, so the tangent is 3x-y-1 = 0

Let P (1,1) be the equation of two tangent lines of the cubic power of curve y = X

(1) If the tangent point is p (1,1), then y '= 3x 2, k = y' (1) = 3,
Therefore, the tangent equation is: Y-1 = 3 (x-1)
That is, 3x-y-2 = 0
(2) The tangent point is (m, m 3), y '(m) = 3M 2 m ≠ 1
That is, the tangent slope k = 3M ~ 2, and then pass through the point P (1,1)
Therefore, k = (m 3 - 1) / (m-1)
That is, 3M 2 = m 2 + m + 1
2m²-m-1=0
(m-1)(2m+1)=0
Because m ≠ 1, so: M = - 1 / 2
Then k = 3M 2 = 3 / 4
Therefore, the tangent equation is: 3x-4y + 1 = 0
To sum up, the two tangent lines are: 3x-y-2 = 0 and 3x-4y + 1 = 0

Find the straight line equation perpendicular to 2x-6y + 1 = 0 and tangent to the cubic power of curve y = x + 3x squared-5

y=x^3+3x^2-5
y'=3x^2+6x
According to the meaning of the title, the tangent slope is the negative reciprocal of the slope of the known straight line, that is - 3
Let 3x ^ 2 + 6x = - 3
The solution is x = - 1
The tangent point is: (- 1, - 3)
So the tangent equation is: 3x + y + 6 = 0

The general formula of the linear equation perpendicular to the straight line 2x-6y + 1 = zero and tangent to the cubic power of the curve y = x + the power 2-1 of 3x is What is the general formula of the linear equation perpendicular to the straight line 2x-6y + 1 = zero and tangent to the third power of the curve y = x + the second power-1 of 3x,

Y = x to the 3rd power + 3x to the 2nd power - 1
Derivation:
y'=3x^2+6x
If the slope of the line 2x-6y + 1 = 0 is 1 / 3, then the slope of the line perpendicular to it is: - 3
So there is: y '= 3x ^ 2 + 6x = - 3
The solution is: x = - 1
Substituting into the curve equation, y = - 1 + 3-1 = 1
The tangent point coordinates are: (- 1,1)
So, the linear equation is: Y-1 = - 3 (x + 1)
That is, y = - 3x-2

The curve represented by the fourth power of equation x - the fourth power of Y - the square of 4x + the square of 4Y = 0 A circle Two parallel lines and a circle Two parallel lines and a circle Two intersecting lines and a circle

Two intersecting lines and a circle are x ^ 4-y ^ 4-4x? + 4Y? = (x? + y?) (x? Y? - 4) - 4 (x? - y?) = (x? Y? (X-Y) (x? 2 + y? - 4) = x + y = 0 or X-Y = 0 or x? + y? = 4? Two intersecting lines and a circle

Find the inflection point and concave convex interval of curve y = 3x4-4x3 + 1

y′=12x3-12x2，
y″=36x2-24x=12x（3x-2）
If y ″ = 0, x = 0 or x = 2
3．
So the inflection points of the curve are (0, 1), (2)
3，11
27）．
When x < 0 or x > 2
At 3, y ″ > 0,
Then the concave interval of the curve is (- ∞, 0), (2
3，+∞），
When 0 < x < 2
At 3, y ″ < 0,
Then the convex interval of the curve is (0, 2
3）．

Finding the inflection point of monotone interval concave convex interval and the knowledge solution of extremum higher number of F (x) = x3-2x2x-1 The number after X is the exponent x to the power of X, followed by x minus 1

First derivative = 3x2-4x + 1 = (3x-1) * (x-1); second derivative = 6x-4;
Monotonically increasing interval (negative infinity, 1 / 3) (1, positive infinity);
Monotone decreasing interval (1 / 3,1);
The inflection point is x = 2 / 3;
Second order greater than zero concave, less than zero convex
The concave interval is (2 / 3, positive infinity);
The convex interval is (negative infinity, 2 / 3);

Ask a question of higher number (about function and limit) If x tends to 0, (1-cosx) ln (1 + x ^ 2) is infinitesimal of higher order than xsinx ^ n, and xsinx ^ n is infinitesimal of higher order than [e ^ (x ^ 2)] - 1? When x tends to 0, what is 1-cosx equivalent to?

N is equal to 2
(1-cosx) ln (1 + x ^ 2) is infinitesimal of higher order than xsinx ^ n
So, n = 2
When x tends to 0, there are some formulas that can be used directly
Answer: when x tends to 0, 1-cosx is equivalent to 1 / 2 (x ^ 2)

Exercises on the limit of functions of higher numbers Find the function f (x) = 1 + X, x > 0, e ^ 1 / x + 1, X

Right limit = 1 + limit (x) = 1 + 0 = 1
When x > 0 -
1 / x > negative infinity
e^1/x -> 0
Left limit = limit (e ^ 1 / x) + 1 = 0 + 1 = 1
So the limit exists and is 1

Why do the left and right limits of a function exist? Why does the limit of a function exist a higher number

The left and right limits are the limits of a function at one point,
It should be noted that the definition of the limit of a function at a point is that there exists a de centered neighborhood. Of course, this neighborhood includes the left and right neighborhoods of this point. It is obvious from the definition that there must be left and right limits for the existence of limit
However, when the limit tends to infinity, there is no left-right limit, because the definition of limit tending to infinity is that there exists a positive or negative number, X is always greater than the positive number or always less than the negative number, so it has nothing to do with all limits

The tangent equation at the point (1,2) of the cubic + 1 Curve y = x is The key is how to find the derivative, how to find y 'what's the formula of derivative ~ ~ urgent