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How to find tangent equation of curve at a certain point according to the geometric meaning of derivative
First, the derivative of the curve is obtained by taking the abscissa of the point into the derivative of the curve. The number obtained is the skew law of the tangent line of the curve at the point. Let the tangent equation be l = KX B, and K is the oblique law, which has been solved earlier. Because the coordinates of the point satisfy the linear equation, B can be obtained by taking the coordinates of the point into the linear equation
What does the derivative of the coordinate equation R (θ) in polar coordinates mean? (for example, the derivative in rectangular coordinates represents the slope of tangent line)
In polar coordinates, the ratio of the polar radius R (θ) to its derivative R '(θ) is equal to the tangent of the angle between the polar radius and the tangent of the curve
Among all the tangents of the curve y = X3 + 3x2 + 6x + 4, the equation of the tangent with the smallest slope is______ .
From the meaning of the title, y '= 3x2 + 6x + 6 = 3 (x2 + 2x) + 6
=3（x+1）2+3，
When x = - 1, y ′ = 3x2 + 6x + 6, the minimum value is 3,
By substituting x = 1 into y = X3 + 3x2 + 6x + 4, y = 14, that is, the coordinates of the tangent point are (1, 14),
The tangent equation is: y-14 = 3 (x-1),
That is, 3x-y + 3 = 0,
So the answer is: 3x-y + 3 = 0
If the slope of tangent is known to be 1, how many tangent equations are there
There are countless
y=x+C
How to find the slope of tangent equation with derivative Tangent equation of curve y = 4x-x ^ 3 at point (- 1, - 3)?
Y 'is the slope of the tangent equation y' = 4-3x ^ 2 = 4-3 * 1 = 1, y = 1 (x + 1) - 3 = X-2
Given the function FX = x ^ 2 * e ^ (- x), when the slope of tangent l of curve y = FX is negative, the value range of intercept of L on X axis is obtained
A:
f(x)=x²×e^(-x)
Derivation:
f'(x)=2xe^(-x)-x²×e^(-x)=x(2-x)e^(-x)
If there is a straight line passing through the point (1,0) and the curve y = x ^ 3 and y = ax ^ 2 + 15 / 4x-9, how to calculate a
Y = x ^ 3, derivative y = 3x ^ 2, straight line and its tangent point are (m, m ^ 3)
Then the straight line passes (m, m ^ 3), (1,0)
The straight line is y = 0 or y = 27 / 4 * (x-1)
If y = 0, then y = ax ^ 2 + 15 / 4x-9 vertex is on the X axis
A = - 25 / 64
If y = 27 / 4 * (x-1), the slope is 27 / 4
The derivative of y = ax ^ 2 + 15 / 4x-9 is y = 2aX + 15 / 4,
The tangent point of a straight line is (n, an ^ 2 + 15 / 4n-9)
2an+15/4=27/4
n=3/(2a)
Straight line crossing (3 / 2,27 / 8), (1,0) (3 / (2a), (63-72a) / 8a)
Launch a = - 1
So a = - 25 / 64 or a = - 1
Finding the extremum and derivative of function Find the extremum of F (x) = 1 / 3x ^ 3 + 4x + 4 The wrong number is the extreme value of F (x) = 1 / 3x ^ 3-4x + 4
f'(x)=x^2-4
By F '(x) = 0, the stable point is obtained, x = 2 or x = - 2
From F '' (x) = 2x
F '' (2) = 4 > 0, so f (x) is a minimum at x = 2, f (2) = 8 / 3-8 + 4 = - 4 / 3
F '' (- 2) = - 4 < 0, so f (x) reaches the maximum at x = - 2, f (- 2) = - 8 / 3 + 8 + 4 = 28 / 3
On the extremum and derivative of function Given the function f (x) = x 3 + ax 2 + BX + C, when x = - 1, the maximum value is 7; when x = 3, the minimum value is obtained. Find the minimum value and the value of a, B, C
(1) Because when x = - 1, f (x) has a maximum value, when x = 3, f (x) has a minimum value. Therefore, if x = - 1 and 3 are substituted into the derivative, and the derivative is equal to 0, two equations about a, B and C can be obtained. According to the maximum value equal to 7, another equation about a, B, C is obtained. The values of a, B, C can be obtained by combining the three equations
(2) Because the function has a minimum value at x = 3, the function value obtained by substituting x = 3 into the original function is the minimum value of the function
（1）∴f（x）=x3+ax2+bx+c
∵f'（x）=3x2+2ax+b
And x = - 1 and x = 3 are extreme points,
So the solution of {f ʹ (- 1) = 3-2a + B = 0f ʹ (3) = 27 + 6A + B = 0 is obtained: a = - 3, B = - 9
F (- 1) = - 1 + A-B + C = - 1-3 + 9 + C = 7, so C = 2
(2) From (1), we know that f (x) = x3-3x2-9x + 2 and x = 3 is its minimum point, so the minimum value of function f (x) is - 25
Solving function extremum with derivative Using derivative to solve the extreme value and pole of y = (m ^ 2 + 1) / (m ^ 2-1) ^ 2. I can't solve the derivative problem with fraction,
For the first derivative of Y, y '= [2m (m ^ 2-1) ^ 2 - (m ^ 2 + 1) 2 (m ^ 2-1) 2m] / [(m ^ 2-1) ^ 4] = 0
(1) When m ^ 2-1 is not equal to 0, M = 0, y = 1, M > 0, y`
RELATED INFORMATIONS
- 1. Exercises on extremum and derivative of function 1. Find the extremum of the following functions （2）y=48x-x^3 （3）y=6+12x-x^3 （4）y=5x+x^3 （5）y=x^2-7x+6 （6）y=x-x^3 5. It is known that f (x) = ax ^ 3 + BX ^ 2 + CX (a ≠ 0) has an extreme value when x = ± 1, and f (1) = - 1, (1) Try to find the values of constants a, B and C (2) Try to judge whether the function gets minimum or maximum when x = ± 1, and explain the reason 6. Let f (x) = x ^ 3-3ax + B (a ≠ 0) (1) If the curve y = f (x) is tangent to the straight line y = 8 at the point (2, f (x)), find the values of a and B; (2) Find the monotone interval and extreme point of function f (x) 7. Let y = f (x) be a cubic function and the image is symmetric about the origin. When x = 1 / 2, the minimum value of F (x) is - 1, and the analytic formula of function f (x) is obtained 8. F (x) = x ^ 3 + 3ax ^ 2 + BX + A ^ 2 has an extreme value of 0 when x = - 1 (1) Find the values of constants A and B; (2) Find: monotone interval of F (x) Don't talk if you don't answer!
- 2. The equation of conic curve. Find the hyperbolic equation with two vertices of ellipse x squared / 16 + y squared / 9 = 1 as focus and ellipse focus as vertex
- 3. The problem of finding tangent by derivative If the function y = f (x) passes through the tangent line of point P (x, f (x)), there are two cases, one is that the point is tangent point, the other is that the point is not tangent point, so there may be two answers My question is, isn't the tangent line only one intersection point with the image of y = f (x)? Since the point P (x, f (x)) is on y = f (x), isn't p the tangent point? If not, isn't the tangent point more than one intersection point with the image of y = f (x)?
- 4. In the triangular vertebrae with o as the vertex, the intersection angle of the three edges passing through O is 30 degrees. Take two points AB on one edge, OA = 4cm, OB = 3cm. At the end point of AB, a rope is used to tightly wrap the side of the triangular pyramid (there is no friction between the rope and the side), and find the shortest rope length between AB and ab
- 5. Can we prove that the line m is perpendicular to the line n?
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