It is known that P (x1, Y1) is any point on the curve y = x ^ 3 + 3x ^ 2 + 4x-10, and the inclination angle of the tangent line at point P is a, so the value range of a is obtained

It is known that P (x1, Y1) is any point on the curve y = x ^ 3 + 3x ^ 2 + 4x-10, and the inclination angle of the tangent line at point P is a, so the value range of a is obtained

y'=3x^2+6x+4
Its range [1, + OO]
Normal slope = - 1 / y '= Tana
tana>=-1
(k-1/4)*pai

Given that point P is any point on the curve y = x ^ 3 + 3x ^ 2 + 4x-10, make the tangent of the curve through the point, and find: (1) the value range of tangent inclination
(2) Tangent equation with minimum slope

1. If y '= 3x & # 178; + 6x + 4 = 3 (x + 1) &# 178; + 1 ≥ 1, then the inclination angle w ∈ [45 ° 90 °)
2. If the minimum slope is k = 1 and x = - 1, then the tangent coordinates are Q (- 1, - 12) and the tangent equation is x-y-11 = 0

Given that the point P is any point on the curve y = x ^ 3 3x ^ 2 4x-10, make the tangent of the curve through the point P
(1) The range of tangent inclination angle α
(2) Tangent equation with minimum slope

y'=3x^2+6x+4=3(x+1)^2+1>=1
The derivative is the tangent slope
So k > = 1
So π / 4