one of the conditions of the inflection point is inflection tangent. what is inflection tangent?


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Question Number 91460 by M±th+et+s last updated on 30/Apr/20

$o n e o f t h e c o n d i t i o n s o f t h e i n f l e c t i o n$ $p o i n t i s i n f l e c t i o n t a n g e n t .$ $w h a t i s i n f l e c t i o n t a n g e n t ?$
	Answered by MJS last updated on 01/May/20
	$it′s the tangent in the inflection point which also intersects the curve in the inflection point I once learned these things with these elementar functions y=ax+b, a≠0 y′=a ⇒ constant slope, no curvature y=ax^2 +bx+c, a≠0 y′=2ax+b zero at x=−(b/(2a)) ⇒ extreme point y′′=2a ⇒ constant curvature depending on the sign of a {: ((a<0 ⇒ maximum)),((a>0 ⇒ minimum)) } at x=−(b/(2a)) y=ax^3 +bx^2 +cx+d, a≠0 y′=3ax^2 +2bx+c zeros at x=−((b±(√(b^2 −3ac)))/(3a)) (1) 2 distinct zeros ∈R ⇒ extreme points (2) 1 double zero ∈R ⇒ no extreme points but an inflection point with horizontal tangent (3) 2 zeros ∉R ⇒ no extreme points in both cases y′′=6ax+2b ⇒ curvature changes zero at x=−(b/(3a)) ⇒ inflection point y′′′=6a { ((a<0 ⇒ curvature changes − to +)),((a>0 ⇒ curvature changes + to −)) :} the inflection tangent also intersects the curve at x=−(b/(3a)) y=ax^4 +bx^3 +cx^2 +dx+e, a≠0 y′=4ax^3 +3bx^2 +2cx+d now it′s getting complicated (1) 3 distinct zeros ∈R (2) 2 double and 1 solitaire zeros ∈R (3) one triple zero ∈R (4) 1 zero ∈R and 2 zeros ∉R we can get 3, 2 or extremes, flat points and a saddle point saddle point: y=x^2 , y′=2x, y′′=2>0 minimum at y=0 y=x^3 , y′=3x^2 , y′′=6x, y′′′=6>0 no extremes inflection point at x=0 with horizontal tangent, curvature changes from − to + y=x^4 , y′=4x^3 , y′′=12x^2 , y′′′=24x minimum at x=0 but y′′=0!? ⇒^? ⇒^? inflection point? but the curvature doesn′t change (y′′′=0) ⇒ this is called a saddle point$
	${it}^{'} s the tangent in the inflection point which$ $also intersects the curve in the inflection$ $point$ $I once learned these things with these$ $elementar functions$ $y = a x + b, a \neq 0$ $y^{'} = a \Rightarrow constant slope, no curvature$ $y = {a x}^{2} + b x + c, a \neq 0$ $y^{'} = 2 a x + b$ $zero at x = - \frac{b}{2 a}$ $\Rightarrow extreme point$ $y^{″} = 2 a \Rightarrow constant curvature depending on$ $the sign of a$ $\begin{matrix} a < 0 \Rightarrow maximum \\ a > 0 \Rightarrow minimum \end{matrix}} at x = - \frac{b}{2 a}$ $y = {a x}^{3} + {b x}^{2} + c x + d, a \neq 0$ $y^{'} = 3 {a x}^{2} + 2 b x + c$ $zeros at x = - \frac{b \pm \sqrt{b^{2} - 3 a c}}{3 a}$ $(1) 2 distinct zeros \in R \Rightarrow extreme points$ $(2) 1 double zero \in R \Rightarrow no extreme points$ $but an inflection point with horizontal$ $tangent$ $(3) 2 zeros \notin R \Rightarrow no extreme points$ $in both cases$ $y^{″} = 6 a x + 2 b \Rightarrow curvature changes$ $zero at x = - \frac{b}{3 a}$ $\Rightarrow inflection point$ $y^{‴} = 6 a {\begin{cases} a < 0 \Rightarrow curvature changes - to + \\ a > 0 \Rightarrow curvature changes + to - \end{cases}$ $the inflection tangent also intersects the$ $curve at x = - \frac{b}{3 a}$ $y = {a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e, a \neq 0$ $y^{'} = 4 {a x}^{3} + 3 {b x}^{2} + 2 c x + d$ $now {it}^{'} s getting complicated$ $(1) 3 distinct zeros \in R$ $(2) 2 double and 1 solitaire zeros \in R$ $(3) one triple zero \in R$ $(4) 1 zero \in R and 2 zeros \notin R$ $we can get 3, 2 or extremes, flat points and$ $a saddle point$ $saddle point :$ $y = x^{2}, y^{'} = 2 x, y^{″} = 2 > 0$ $minimum at y = 0$ $y = x^{3}, y^{'} = 3 x^{2}, y^{″} = 6 x, y^{‴} = 6 > 0$ $no extremes$ $inflection point at x = 0 with horizontal$ $tangent, curvature changes from - to +$ $y = x^{4}, y^{'} = 4 x^{3}, y^{″} = 12 x^{2}, y^{‴} = 24 x$ $minimum at x = 0 but y^{″} = 0! ? \overset{?}{\Rightarrow}$ $\overset{?}{\Rightarrow} inflection point ? but the curvature$ ${doesn}^{'} t change (y^{‴} = 0) \Rightarrow this is called$ $a saddle point$
	Commented by M±th+et+s last updated on 01/May/20

	$v e r y c o o l e x p l a n a t i o n . t h a n k y o u v e r y m u c h$
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