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Advanced Mathematics (Seventh Edition) Tongji University exercises 3-4 personal solutions (the last 8 questions)

2022-07-28 03:43:00 【Navigator_ Z】

Advanced mathematics （ The seventh edition ） Tongji University exercises 3-4（ after 8 topic ）

 

9.   Determine the concavity and convexity of the following curves ： \begin{aligned}&9. \ Determine the concavity and convexity of the following curves ：&\end{aligned} ​9.  Determine the concavity and convexity of the following curves ：​​

   ( 1 )    y = 4 x − x 2 ;                                         ( 2 )    y = s h   x ;    ( 3 )    y = x + 1 x   ( x > 0 ) ;                              ( 4 )    y = x a r c t a n   x . \begin{aligned} &\ \ (1)\ \ y=4x-x^2;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=sh\ x;\\\\ &\ \ (3)\ \ y=x+\frac{1}{x}\ (x \gt 0);\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ y=xarctan\ x. & \end{aligned} ​  (1)  y=4x−x2;                                        (2)  y=sh x;  (3)  y=x+x1​ (x>0);                             (4)  y=xarctan x.​​

Explain ：

   ( 1 )   y ′ = 4 − 2 x , y ′ ′ = − 2 < 0 , So the curve y = 4 x − x 2 stay ( − ∞ ,   + ∞ ) The inside is convex .    ( 2 )   y ′ = c h   x , y ′ ′ = s h   x , Make y ′ ′ = 0 , have to x = 0 .          When − ∞ < x < 0 when , y ′ ′ < 0 , curve y = s h   x stay ( − ∞ ,   0 ] It's convex .          When 0 < x < + ∞ when , y ′ ′ > 0 , curve y = s h   x stay [ 0 ,   + ∞ ) The top is concave .    ( 3 )   y ′ = 1 − 1 x 2 , y ′ ′ = 2 x 3 > 0 （ x > 0 ）, So the curve y = x + 1 x stay ( 0 ,   + ∞ ) The inside is concave .    ( 4 )   y ′ = a r c t a n   x + x 1 + x 2 , y ′ ′ = 1 1 + x 2 + 1 + x 2 − x ⋅ 2 x ( 1 + x 2 ) 2 = 2 ( 1 + x 2 ) 2 > 0 ,          So the curve y = x a r c t a n   x stay ( − ∞ ,   + ∞ ) The inside is concave . \begin{aligned} &\ \ (1)\ y'=4-2x,y''=-2 \lt 0, So the curve y=4x-x^2 stay (-\infty, \ +\infty) The inside is convex .\\\\ &\ \ (2)\ y'=ch\ x,y''=sh\ x, Make y''=0, have to x=0.\\\\ &\ \ \ \ \ \ \ \ When -\infty \lt x \lt 0 when ,y'' \lt 0, curve y=sh\ x stay (-\infty, \ 0] It's convex .\\\\ &\ \ \ \ \ \ \ \ When 0 \lt x \lt +\infty when ,y'' \gt 0, curve y=sh\ x stay [0, \ +\infty) The top is concave .\\\\ &\ \ (3)\ y'=1-\frac{1}{x^2},y''=\frac{2}{x^3} \gt 0（x \gt 0）, So the curve y=x+\frac{1}{x} stay (0, \ +\infty) The inside is concave .\\\\ &\ \ (4)\ y'=arctan\ x+\frac{x}{1+x^2},y''=\frac{1}{1+x^2}+\frac{1+x^2-x\cdot 2x}{(1+x^2)^2}=\frac{2}{(1+x^2)^2} \gt 0,\\\\ &\ \ \ \ \ \ \ \ So the curve y=xarctan\ x stay (-\infty, \ +\infty) The inside is concave . & \end{aligned} ​  (1) y′=4−2x,y′′=−2<0, So the curve y=4x−x2 stay (−∞, +∞) The inside is convex .  (2) y′=ch x,y′′=sh x, Make y′′=0, have to x=0.         When −∞<x<0 when ,y′′<0, curve y=sh x stay (−∞, 0] It's convex .         When 0<x<+∞ when ,y′′>0, curve y=sh x stay [0, +∞) The top is concave .  (3) y′=1−x21​,y′′=x32​>0（x>0）, So the curve y=x+x1​ stay (0, +∞) The inside is concave .  (4) y′=arctan x+1+x2x​,y′′=1+x21​+(1+x2)21+x2−x⋅2x​=(1+x2)22​>0,         So the curve y=xarctan x stay (−∞, +∞) The inside is concave .​​

10.   Find the inflection point and concave or convex interval of the following function graph ： \begin{aligned}&10. \ Find the inflection point and concave or convex interval of the following function graph ：&\end{aligned} ​10.  Find the inflection point and concave or convex interval of the following function graph ：​​

   ( 1 )    y = x 3 − 5 x 2 + 3 x + 5 ;                       ( 2 )    y = x e − x ;    ( 3 )    y = ( x + 1 ) 4 + e x ;                               ( 4 )    y = l n ( x 2 + 1 ) ;    ( 5 )    y = e a r c t a n   x ;                                         ( 6 )    y = x 4 ( 12 l n   x − 7 ) . \begin{aligned} &\ \ (1)\ \ y=x^3-5x^2+3x+5;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ y=xe^{-x};\\\\ &\ \ (3)\ \ y=(x+1)^4+e^x;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ y=ln(x^2+1);\\\\ &\ \ (5)\ \ y=e^{arctan\ x};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ y=x^4(12ln\ x-7). & \end{aligned} ​  (1)  y=x3−5x2+3x+5;                      (2)  y=xe−x;  (3)  y=(x+1)4+ex;                              (4)  y=ln(x2+1);  (5)  y=earctan x;                                        (6)  y=x4(12ln x−7).​​

Explain ：

   ( 1 )   y ′ = 3 x 2 − 10 x + 3 , y ′ ′ = 6 x − 10 , Make y ′ ′ = 0 , have to x = 5 3 , y = 20 27 .          When − ∞ < x < 5 3 when , y ′ ′ < 0 , So the curve is ( − ∞ ,   5 3 ] It's convex ,          When 5 3 < x < + ∞ when , y ′ ′ > 0 , So the curve is [ 5 3 ,   + ∞ ) The top is concave , therefore , spot ( 5 3 ,   20 27 ) It's the inflection point .    ( 2 )   y ′ = e − x − x e − x = ( 1 − x ) e − x , y ′ ′ = − e − x + ( 1 − x ) ( − e − x ) = e − x ( x − 2 ) .          Make y ′ ′ = 0 , have to x = 2 , y = 2 e 2 .          When − ∞ < x < 2 when , y ′ ′ < 0 , So the curve is ( − ∞ ,   2 ] It's convex ,          When 2 < x < + ∞ when , y ′ ′ > 0 , So the curve is ( 2 ,   + ∞ ) The top is concave , therefore , spot ( 2 ,   2 e 2 ) It's the inflection point .    ( 3 )   y ′ = 4 ( x + 1 ) 3 + e x , y ′ ′ = 12 ( x + 1 ) 2 + e x > 0 , So the curve is ( − ∞ ,   + ∞ ) The inside is concave , The curve has no inflection point .    ( 4 )   y ′ = 2 x x 2 + 1 , y ′ ′ = 2 ( x 2 + 1 ) − 2 x ⋅ 2 x ( x 2 + 1 ) 2 = − 2 ( x − 1 ) ( x + 1 ) ( x 2 + 1 ) 2 . Make y ′ ′ = 0 , have to x 1 = − 1 , x 2 = 1.          When − ∞ < x < − 1 when , y ′ ′ < 0 , So the curve is ( − ∞ ,   − 1 ] It's convex ,          When − 1 < x < 1 when , y ′ ′ > 0 , So the curve is [ − 1 ,   1 ] The top is concave ,          When 1 < x   + ∞ when , y ′ ′ < 0 , So the curve is [ 1 ,   + ∞ ) It's convex . The curve has two inflection points , The difference is ( − 1 ,   l n   2 ) Sum point ( 1 ,   l n   2 ) .    ( 5 )   y ′ = e a r c t a n   x 1 + x 2 , y ′ ′ = − 2 e a r c t a n   x ( x − 1 2 ) ( 1 + x 2 ) 2 , Make y ′ ′ = 0 , have to x = 1 2 , y = e a r c t a n   1 2 ,          When − ∞ < x < 1 2 when , y ′ ′ > 0 , So the curve is ( − ∞ ,   1 2 ] The top is concave ,          When 1 2 < x < + ∞ when , y ′ ′ < 0 , So the curve is [ 1 2 ,   + ∞ ) Convex . So point ( 1 2 ,   e a r c t a n   1 2 ) It's the inflection point .    ( 6 )   y ′ = 4 x 3 ( 12 l n   x − 7 ) + x 4 ⋅ 12 1 x = 4 x 3 ( 12 l n   x − 4 ) , y ′ ′ = 12 x 2 ( 12 l n   x − 4 ) + 4 x 3 ⋅ 12 1 x = 144 x 2 l n   x   ( x > 0 ) .          Make y ′ ′ = 0 , have to x = 1 , y = − 7 . When 0 < x < 1 when , y ′ ′ < 0 , So the curve is ( 0 ,   1 ] It's convex ,          When 1 < x < + ∞ when , y ′ ′ > 0 , So the curve is [ 1 ,   + ∞ ) The top is concave , So point ( 1 ,   − 7 ) It's the inflection point . \begin{aligned} &\ \ (1)\ y'=3x^2-10x+3,y''=6x-10, Make y''=0, have to x=\frac{5}{3},y=\frac{20}{27}.\\\\ &\ \ \ \ \ \ \ \ When -\infty \lt x \lt \frac{5}{3} when ,y'' \lt 0, So the curve is \left(-\infty, \ \frac{5}{3}\right] It's convex ,\\\\ &\ \ \ \ \ \ \ \ When \frac{5}{3} \lt x \lt +\infty when ,y'' \gt 0, So the curve is \left[\frac{5}{3}, \ +\infty\right) The top is concave , therefore , spot \left(\frac{5}{3}, \ \frac{20}{27}\right) It's the inflection point .\\\\ &\ \ (2)\ y'=e^{-x}-xe^{-x}=(1-x)e^{-x},y''=-e^{-x}+(1-x)(-e^{-x})=e^{-x}(x-2).\\\\ &\ \ \ \ \ \ \ \ Make y''=0, have to x=2,y=\frac{2}{e^2}.\\\\ &\ \ \ \ \ \ \ \ When -\infty \lt x \lt 2 when ,y'' \lt 0, So the curve is (-\infty, \ 2] It's convex ,\\\\ &\ \ \ \ \ \ \ \ When 2 \lt x \lt +\infty when ,y'' \gt 0, So the curve is (2, \ +\infty) The top is concave , therefore , spot \left(2, \ \frac{2}{e^2}\right) It's the inflection point .\\\\ &\ \ (3)\ y'=4(x+1)^3+e^x,y''=12(x+1)^2+e^x \gt 0, So the curve is (-\infty, \ +\infty) The inside is concave , The curve has no inflection point .\\\\ &\ \ (4)\ y'=\frac{2x}{x^2+1},y''=\frac{2(x^2+1)-2x \cdot 2x}{(x^2+1)^2}=\frac{-2(x-1)(x+1)}{(x^2+1)^2}. Make y''=0, have to x_1=-1,x_2=1.\\\\ &\ \ \ \ \ \ \ \ When -\infty \lt x \lt -1 when ,y'' \lt 0, So the curve is (-\infty, \ -1] It's convex ,\\\\ &\ \ \ \ \ \ \ \ When -1 \lt x \lt 1 when ,y'' \gt 0, So the curve is [-1, \ 1] The top is concave ,\\\\ &\ \ \ \ \ \ \ \ When 1 \lt x \ +\infty when ,y'' \lt 0, So the curve is [1, \ +\infty) It's convex . The curve has two inflection points , The difference is (-1, \ ln\ 2) Sum point (1, \ ln\ 2).\\\\ &\ \ (5)\ y'=\frac{e^{arctan\ x}}{1+x^2},y''=\frac{-2e^{arctan\ x}\left(x-\frac{1}{2}\right)}{(1+x^2)^2}, Make y''=0, have to x=\frac{1}{2},y=e^{arctan\ \frac{1}{2}},\\\\ &\ \ \ \ \ \ \ \ When -\infty \lt x \lt \frac{1}{2} when ,y'' \gt 0, So the curve is \left(-\infty, \ \frac{1}{2}\right] The top is concave ,\\\\ &\ \ \ \ \ \ \ \ When \frac{1}{2} \lt x \lt +\infty when ,y'' \lt 0, So the curve is \left[\frac{1}{2}, \ +\infty\right) Convex . So point \left(\frac{1}{2}, \ e^{arctan\ \frac{1}{2}}\right) It's the inflection point .\\\\ &\ \ (6)\ y'=4x^3(12ln\ x-7)+x^4 \cdot 12\frac{1}{x}=4x^3(12ln\ x-4),y''=12x^2(12ln\ x-4)+4x^3\cdot 12\frac{1}{x}=144x^2ln\ x\ (x \gt 0).\\\\ &\ \ \ \ \ \ \ \ Make y''=0, have to x=1,y=-7. When 0 \lt x \lt 1 when ,y'' \lt 0, So the curve is (0, \ 1] It's convex ,\\\\ &\ \ \ \ \ \ \ \ When 1 \lt x \lt +\infty when ,y'' \gt 0, So the curve is [1, \ +\infty) The top is concave , So point (1, \ -7) It's the inflection point . & \end{aligned} ​  (1) y′=3x2−10x+3,y′′=6x−10, Make y′′=0, have to x=35​,y=2720​.         When −∞<x<35​ when ,y′′<0, So the curve is (−∞, 35​] It's convex ,         When 35​<x<+∞ when ,y′′>0, So the curve is [35​, +∞) The top is concave , therefore , spot (35​, 2720​) It's the inflection point .  (2) y′=e−x−xe−x=(1−x)e−x,y′′=−e−x+(1−x)(−e−x)=e−x(x−2).         Make y′′=0, have to x=2,y=e22​.         When −∞<x<2 when ,y′′<0, So the curve is (−∞, 2] It's convex ,         When 2<x<+∞ when ,y′′>0, So the curve is (2, +∞) The top is concave , therefore , spot (2, e22​) It's the inflection point .  (3) y′=4(x+1)3+ex,y′′=12(x+1)2+ex>0, So the curve is (−∞, +∞) The inside is concave , The curve has no inflection point .  (4) y′=x2+12x​,y′′=(x2+1)22(x2+1)−2x⋅2x​=(x2+1)2−2(x−1)(x+1)​. Make y′′=0, have to x1​=−1,x2​=1.         When −∞<x<−1 when ,y′′<0, So the curve is (−∞, −1] It's convex ,         When −1<x<1 when ,y′′>0, So the curve is [−1, 1] The top is concave ,         When 1<x +∞ when ,y′′<0, So the curve is [1, +∞) It's convex . The curve has two inflection points , The difference is (−1, ln 2) Sum point (1, ln 2).  (5) y′=1+x2earctan x​,y′′=(1+x2)2−2earctan x(x−21​)​, Make y′′=0, have to x=21​,y=earctan 21​,         When −∞<x<21​ when ,y′′>0, So the curve is (−∞, 21​] The top is concave ,         When 21​<x<+∞ when ,y′′<0, So the curve is [21​, +∞) Convex . So point (21​, earctan 21​) It's the inflection point .  (6) y′=4x3(12ln x−7)+x4⋅12x1​=4x3(12ln x−4),y′′=12x2(12ln x−4)+4x3⋅12x1​=144x2ln x (x>0).         Make y′′=0, have to x=1,y=−7. When 0<x<1 when ,y′′<0, So the curve is (0, 1] It's convex ,         When 1<x<+∞ when ,y′′>0, So the curve is [1, +∞) The top is concave , So point (1, −7) It's the inflection point .​​

11.   Using the concavity and convexity of function graph , Prove the following inequality ： \begin{aligned}&11. \ Using the concavity and convexity of function graph , Prove the following inequality ：&\end{aligned} ​11.  Using the concavity and convexity of function graph , Prove the following inequality ：​​

   ( 1 )    1 2 ( x n + y n ) > ( x + y 2 ) n   ( x > 0 , y > 0 , x ≠ y , n > 1 ) ;    ( 2 )    e x + e y 2 > e x + y 2   ( x ≠ y ) ;    ( 3 )    x l n   x + y l n   y > ( x + y ) l n   x + y 2   ( x > 0 , y > 0 , x ≠ y ) . \begin{aligned} &\ \ (1)\ \ \frac{1}{2}(x^n+y^n) \gt \left(\frac{x+y}{2}\right)^n\ (x \gt 0,y \gt 0,x \neq y,n \gt 1);\\\\ &\ \ (2)\ \ \frac{e^x+e^y}{2} \gt e^{\frac{x+y}{2}}\ (x \neq y);\\\\ &\ \ (3)\ \ xln\ x+yln\ y \gt (x+y)ln\ \frac{x+y}{2}\ (x \gt 0,y \gt 0,x \neq y). & \end{aligned} ​  (1)  21​(xn+yn)>(2x+y​)n (x>0,y>0,x=y,n>1);  (2)  2ex+ey​>e2x+y​ (x=y);  (3)  xln x+yln y>(x+y)ln 2x+y​ (x>0,y>0,x=y).​​

Explain ：

   ( 1 )   take f ( t ) = t n , t ∈ ( 0 ,   + ∞ ) . f ′ ( t ) = n t n − 1 , f ′ ′ ( t ) = n ( n − 1 ) t n − 2 , t ∈ ( 0 ,   + ∞ ) .          When n > 1 when , f ′ ′ ( t ) > 0 , t ∈ ( 0 ,   + ∞ ) . therefore f ( t ) stay ( 0 ,   + ∞ ) The inner figure is concave , To any x > 0 , y > 0 , x ≠ y , There is always          1 2 [ f ( x ) + f ( y ) ] > f ( x + y 2 ) , namely 1 2 ( x n + y n ) > ( x + y 2 ) n   ( x > 0 , y > 0 , x ≠ y , n > 1 ) .    ( 2 )   take f ( t ) = e t , t ∈ ( − ∞ ,   + ∞ ) , f ′ ( t ) = e t , f ′ ′ ( t ) = e t > 0 , t ∈ ( − ∞ ,   + ∞ ) .          therefore f ( t ) stay ( − ∞ ,   + ∞ ) The inner figure is concave , To any x , y ∈ ( − ∞ ,   + ∞ ) , x ≠ y , There is always          1 2 [ f ( x ) + f ( y ) ] > f ( x + y 2 ) , namely e x + e y 2 > e x + y 2   ( x ≠ y )    ( 3 )   take f ( t ) = t l n   t , t ∈ ( 0 ,   + ∞ ) , f ′ ( t ) = l n   t + 1 , f ′ ′ ( t ) = 1 t > 0 , t ∈ ( 0 ,   + ∞ ) ,          therefore f ( t ) stay ( 0 ,   + ∞ ) The inner figure is concave , To any x , y ∈ ( 0 ,   + ∞ ) , x ≠ y , There is always          1 2 [ f ( x ) + f ( y ) ] > f ( x + y 2 ) , namely 1 2 ( x l n   x + y l n   y ) > x + y 2 l n x + y 2 ,           x l n   x + y l n   y > ( x + y ) l n   x + y 2   ( x > 0 , y > 0 , x ≠ y ) \begin{aligned} &\ \ (1)\ take f(t)=t^n,t \in (0, \ +\infty).f'(t)=nt^{n-1},f''(t)=n(n-1)t^{n-2},t \in (0, \ +\infty).\\\\ &\ \ \ \ \ \ \ \ When n \gt 1 when ,f''(t) \gt 0,t \in (0, \ +\infty). therefore f(t) stay (0, \ +\infty) The inner figure is concave , To any x \gt 0,y \gt 0,x \neq y, There is always \\\\ &\ \ \ \ \ \ \ \ \frac{1}{2}[f(x)+f(y)] \gt f\left(\frac{x+y}{2}\right), namely \frac{1}{2}(x^n+y^n) \gt \left(\frac{x+y}{2}\right)^n\ (x \gt 0,y \gt 0,x \neq y,n \gt 1).\\\\ &\ \ (2)\ take f(t)=e^t,t \in (-\infty, \ +\infty),f'(t)=e^t,f''(t)=e^t \gt 0,t \in (-\infty, \ +\infty).\\\\ &\ \ \ \ \ \ \ \ therefore f(t) stay (-\infty, \ +\infty) The inner figure is concave , To any x, y \in (-\infty, \ +\infty),x \neq y, There is always \\\\ &\ \ \ \ \ \ \ \ \frac{1}{2}[f(x)+f(y)] \gt f\left(\frac{x+y}{2}\right), namely \frac{e^x+e^y}{2} \gt e^{\frac{x+y}{2}}\ (x \neq y)\\\\ &\ \ (3)\ take f(t)=tln\ t,t \in (0, \ +\infty),f'(t)=ln\ t+1,f''(t)=\frac{1}{t} \gt 0,t \in (0, \ +\infty),\\\\ &\ \ \ \ \ \ \ \ therefore f(t) stay (0, \ +\infty) The inner figure is concave , To any x, y \in (0, \ +\infty),x \neq y, There is always \\\\ &\ \ \ \ \ \ \ \ \frac{1}{2}[f(x)+f(y)] \gt f\left(\frac{x+y}{2}\right), namely \frac{1}{2}(xln\ x+yln\ y) \gt \frac{x+y}{2}ln\frac{x+y}{2},\\\\ &\ \ \ \ \ \ \ \ \ xln\ x+yln\ y \gt (x+y)ln\ \frac{x+y}{2}\ (x \gt 0,y \gt 0,x \neq y) & \end{aligned} ​  (1)  take f(t)=tn,t∈(0, +∞).f′(t)=ntn−1,f′′(t)=n(n−1)tn−2,t∈(0, +∞).         When n>1 when ,f′′(t)>0,t∈(0, +∞). therefore f(t) stay (0, +∞) The inner figure is concave , To any x>0,y>0,x=y, There is always         21​[f(x)+f(y)]>f(2x+y​), namely 21​(xn+yn)>(2x+y​)n (x>0,y>0,x=y,n>1).  (2)  take f(t)=et,t∈(−∞, +∞),f′(t)=et,f′′(t)=et>0,t∈(−∞, +∞).         therefore f(t) stay (−∞, +∞) The inner figure is concave , To any x,y∈(−∞, +∞),x=y, There is always         21​[f(x)+f(y)]>f(2x+y​), namely 2ex+ey​>e2x+y​ (x=y)  (3)  take f(t)=tln t,t∈(0, +∞),f′(t)=ln t+1,f′′(t)=t1​>0,t∈(0, +∞),         therefore f(t) stay (0, +∞) The inner figure is concave , To any x,y∈(0, +∞),x=y, There is always         21​[f(x)+f(y)]>f(2x+y​), namely 21​(xln x+yln y)>2x+y​ln2x+y​,         xln x+yln y>(x+y)ln 2x+y​ (x>0,y>0,x=y)​​

12.   Try to prove the curve y = x − 1 x 2 + 1 There are three inflection points on the same straight line . \begin{aligned}&12. \ Try to prove the curve y=\frac{x-1}{x^2+1} There are three inflection points on the same straight line .&\end{aligned} ​12.  Try to prove the curve y=x2+1x−1​ There are three inflection points on the same straight line .​​

Explain ：

   y ′ = ( x 2 + 1 ) − 2 x ( x − 1 ) ( x 2 + 1 ) 2 = − x 2 + 2 x + 1 ( x 2 + 1 ) 2 ,    y ′ ′ = ( − 2 x + 2 ) ( x 2 + 1 ) 2 − 2 ( x 2 + 1 ) ⋅ 2 x ( − x 2 + 2 x + 1 ) ( x 2 + 1 ) 4 = 2 x 3 − 6 x 2 − 6 x + 2 ( x 2 + 1 ) 3 = 2 ( x + 1 ) [ x − ( 2 − 3 ) ] [ x − ( 2 + 3 ) ] ( x 2 + 1 ) 3    Make y ′ ′ = 0 , have to x 1 = − 1 , x 2 = 2 − 3 , x 3 = 2 + 3 , y 1 = − 1 , y 2 = 1 − 3 4 ( 2 − 3 ) , y 3 = 1 + 3 4 ( 2 + 3 )    When − ∞ < x < − 1 when , y ′ ′ < 0 , So the curve is ( − ∞ ,   − 1 ] It's convex ,    When − 1 < x < 2 − 3 when , y ′ ′ > 0 , So the curve is [ − 1 ,   2 − 3 ] The top is concave ,    When 2 − 3 < x < 2 + 3 when , y ′ ′ < 0 , So the curve is [ 2 − 3 ,   2 + 3 ] It's convex ,    When 2 + 3 < x < + ∞ when , y ′ ′ > 0 , So the curve is [ 2 + 3 ,   + ∞ ) The top is concave ,    So the curve has three inflection points , by ( − 1 ,   − 1 ) , ( 2 − 3 ,   1 − 3 4 ( 2 − 3 ) ) , ( 2 + 3 ,   1 + 3 4 ( 2 + 3 ) )    because 1 − 3 4 ( 2 − 3 ) − ( − 1 ) 2 − 3 − ( − 1 ) = 1 + 3 4 ( 2 + 3 ) − ( − 1 ) 2 + 3 − ( − 1 ) = 1 4 , So the three inflection points are in a straight line . \begin{aligned} &\ \ y'=\frac{(x^2+1)-2x(x-1)}{(x^2+1)^2}=\frac{-x^2+2x+1}{(x^2+1)^2},\\\\ &\ \ y''=\frac{(-2x+2)(x^2+1)^2-2(x^2+1)\cdot 2x(-x^2+2x+1)}{(x^2+1)^4}=\frac{2x^3-6x^2-6x+2}{(x^2+1)^3}=\frac{2(x+1)[x-(2-\sqrt{3})][x-(2+\sqrt{3})]}{(x^2+1)^3}\\\\ &\ \ Make y''=0, have to x_1=-1,x_2=2-\sqrt{3},x_3=2+\sqrt{3},y_1=-1,y_2=\frac{1-\sqrt{3}}{4(2-\sqrt{3})},y_3=\frac{1+\sqrt{3}}{4(2+\sqrt{3})}\\\\ &\ \ When -\infty \lt x \lt -1 when ,y'' \lt 0, So the curve is (-\infty, \ -1] It's convex ,\\\\ &\ \ When -1 \lt x \lt 2-\sqrt{3} when ,y'' \gt 0, So the curve is [-1, \ 2-\sqrt{3}] The top is concave ,\\\\ &\ \ When 2-\sqrt{3} \lt x \lt 2+\sqrt{3} when ,y'' \lt 0, So the curve is [2-\sqrt{3}, \ 2+\sqrt{3}] It's convex ,\\\\ &\ \ When 2+\sqrt{3} \lt x \lt +\infty when ,y'' \gt 0, So the curve is [2+\sqrt{3}, \ +\infty) The top is concave ,\\\\ &\ \ So the curve has three inflection points , by (-1, \ -1),\left(2-\sqrt{3}, \ \frac{1-\sqrt{3}}{4(2-\sqrt{3})}\right),\left(2+\sqrt{3}, \ \frac{1+\sqrt{3}}{4(2+\sqrt{3})}\right)\\\\ &\ \ because \frac{\frac{1-\sqrt{3}}{4(2-\sqrt{3})}-(-1)}{2-\sqrt{3}-(-1)}=\frac{\frac{1+\sqrt{3}}{4(2+\sqrt{3})}-(-1)}{2+\sqrt{3}-(-1)}=\frac{1}{4}, So the three inflection points are in a straight line . & \end{aligned} ​  y′=(x2+1)2(x2+1)−2x(x−1)​=(x2+1)2−x2+2x+1​,  y′′=(x2+1)4(−2x+2)(x2+1)2−2(x2+1)⋅2x(−x2+2x+1)​=(x2+1)32x3−6x2−6x+2​=(x2+1)32(x+1)[x−(2−3​)][x−(2+3​)]​   Make y′′=0, have to x1​=−1,x2​=2−3​,x3​=2+3​,y1​=−1,y2​=4(2−3​)1−3​​,y3​=4(2+3​)1+3​​   When −∞<x<−1 when ,y′′<0, So the curve is (−∞, −1] It's convex ,   When −1<x<2−3​ when ,y′′>0, So the curve is [−1, 2−3​] The top is concave ,   When 2−3​<x<2+3​ when ,y′′<0, So the curve is [2−3​, 2+3​] It's convex ,   When 2+3​<x<+∞ when ,y′′>0, So the curve is [2+3​, +∞) The top is concave ,   So the curve has three inflection points , by (−1, −1),(2−3​, 4(2−3​)1−3​​),(2+3​, 4(2+3​)1+3​​)   because 2−3​−(−1)4(2−3​)1−3​​−(−1)​=2+3​−(−1)4(2+3​)1+3​​−(−1)​=41​, So the three inflection points are in a straight line .​​

13.   ask a 、 b Why is it worth it , spot ( 1 ,   3 ) Is a curve y = a x 3 + b x 2 Inflection point . \begin{aligned}&13. \ ask a、b Why is it worth it , spot (1, \ 3) Is a curve y=ax^3+bx^2 Inflection point .&\end{aligned} ​13.  ask a、b Why is it worth it , spot (1, 3) Is a curve y=ax3+bx2 Inflection point .​​

Explain ：

   y ′ = 3 a x 2 + 2 b x , y ′ ′ = 6 a x + 2 b = 6 a ( x + b 3 a ) , Make y ′ ′ = 0 , have to x = − b 3 a ,    When − ∞ < x < − b 3 a when , y ′ ′ < 0 , So the curve is ( − ∞ ,   − b 3 a ] It's convex ,    When − b 3 a < x < + ∞ when , y ′ ′ > 0 , So the curve is [ − b 3 a ,   + ∞ ) The top is concave ,    When x = − b 3 a when , y = a ( − b 3 a ) 3 + b ( − b 3 a ) 2 = 2 b 3 27 a 2 . because y ′ ′ stay x Different numbers on both sides of , So point ( − b 3 a ,   2 b 3 27 a 2 ) Is the inflection point of the curve .    Make some ( 1 ,   3 ) It's the inflection point , be − b 3 a = 1 , 2 b 3 27 a 2 = 3 , Solution a = − 3 2 , b = 9 2 . \begin{aligned} &\ \ y'=3ax^2+2bx,y''=6ax+2b=6a\left(x+\frac{b}{3a}\right), Make y''=0, have to x=-\frac{b}{3a},\\\\ &\ \ When -\infty \lt x \lt -\frac{b}{3a} when ,y'' \lt 0, So the curve is \left(-\infty, \ -\frac{b}{3a}\right] It's convex ,\\\\ &\ \ When -\frac{b}{3a} \lt x \lt +\infty when ,y'' \gt 0, So the curve is \left[-\frac{b}{3a}, \ +\infty\right) The top is concave ,\\\\ &\ \ When x=-\frac{b}{3a} when ,y=a\left(-\frac{b}{3a}\right)^3+b\left(-\frac{b}{3a}\right)^2=\frac{2b^3}{27a^2}. because y'' stay x Different numbers on both sides of , So point \left(-\frac{b}{3a}, \ \frac{2b^3}{27a^2}\right) Is the inflection point of the curve .\\\\ &\ \ Make some (1, \ 3) It's the inflection point , be -\frac{b}{3a}=1,\frac{2b^3}{27a^2}=3, Solution a=-\frac{3}{2},b=\frac{9}{2}. & \end{aligned} ​  y′=3ax2+2bx,y′′=6ax+2b=6a(x+3ab​), Make y′′=0, have to x=−3ab​,   When −∞<x<−3ab​ when ,y′′<0, So the curve is (−∞, −3ab​] It's convex ,   When −3ab​<x<+∞ when ,y′′>0, So the curve is [−3ab​, +∞) The top is concave ,   When x=−3ab​ when ,y=a(−3ab​)3+b(−3ab​)2=27a22b3​. because y′′ stay x Different numbers on both sides of , So point (−3ab​, 27a22b3​) Is the inflection point of the curve .   Make some (1, 3) It's the inflection point , be −3ab​=1,27a22b3​=3, Solution a=−23​,b=29​.​​

14.   Try to determine the curve y = a x 3 + b x 2 + c x + d Medium a 、 b 、 c 、 d , bring x = − 2 The curve at has a horizontal tangent , ( 1 ,   − 10 ) It's the inflection point ,        And the point ( − 2 ,   44 ) On the curve . \begin{aligned}&14. \ Try to determine the curve y=ax^3+bx^2+cx+d Medium a、b、c、d, bring x=-2 The curve at has a horizontal tangent ,(1, \ -10) It's the inflection point ,\\\\&\ \ \ \ \ \ And the point (-2, \ 44) On the curve .&\end{aligned} ​14.  Try to determine the curve y=ax3+bx2+cx+d Medium a、b、c、d, bring x=−2 The curve at has a horizontal tangent ,(1, −10) It's the inflection point ,       And the point (−2, 44) On the curve .​​

Explain ：

   y ′ = 3 a x 2 + 2 b x + c , y ′ ′ = 6 a x + 2 b , According to the meaning , Yes y ( − 2 ) = 44 , y ′ ( − 2 ) = 0 , y ( 1 ) = 10 , y ′ ′ ( 1 ) = 0 ,    namely { − 8 a + 4 b − 2 c + d = 44 , 12 a − 4 b + c = 0 , a + b + c + d = − 10 , 6 a + 2 b = 0.    To solve the equation a = 1 , b = − 3 , c = − 24 , d = 16. \begin{aligned} &\ \ y'=3ax^2+2bx+c,y''=6ax+2b, According to the meaning , Yes y(-2)=44,y'(-2)=0,y(1)=10,y''(1)=0,\\\\ &\ \ namely \begin{cases}-8a+4b-2c+d=44,\\\\12a-4b+c=0,\\\\a+b+c+d=-10,\\\\6a+2b=0.\end{cases}\\\\ &\ \ To solve the equation a=1,b=-3,c=-24,d=16. & \end{aligned} ​  y′=3ax2+2bx+c,y′′=6ax+2b, According to the meaning , Yes y(−2)=44,y′(−2)=0,y(1)=10,y′′(1)=0,   namely ⎩⎨⎧​−8a+4b−2c+d=44,12a−4b+c=0,a+b+c+d=−10,6a+2b=0.​   To solve the equation a=1,b=−3,c=−24,d=16.​​

15.   Try to decide y = k ( x 2 − 3 ) 2 in k Value , Make the normal at the inflection point of the curve pass through the origin . \begin{aligned}&15. \ Try to decide y=k(x^2-3)^2 in k Value , Make the normal at the inflection point of the curve pass through the origin .&\end{aligned} ​15.  Try to decide y=k(x2−3)2 in k Value , Make the normal at the inflection point of the curve pass through the origin .​​

Explain ：

   y ′ = 2 k ( x 2 − 3 ) ⋅ 2 x = 4 k x ( x 2 − 3 ) , y ′ ′ = 4 k ( x 2 − 3 ) + 4 k x ⋅ 2 x = 12 k x ( x − 1 ) ( x + 1 ) .    Make y ′ ′ = 0 , have to x 1 = − 1 , x 2 = 1 , y 1 = 4 k , y 2 = 4 k .    When − ∞ < x < − 1 when , y ′ ′ > 0 , So the curve is ( − ∞ ,   − 1 ] The top is concave ,    When − 1 < x < 1 when , y ′ ′ < 0 , So the curve is [ − 1 ,   1 ] It's convex ,    When 1 < x < + ∞ when , y ′ ′ > 0 , So the curve is [ 1 ,   + ∞ ) Concave up , So point ( − 1 ,   4 k ) and ( 1 ,   4 k ) Is the inflection point of the curve .    When x = − 1 when , y ′ = 8 k , So a little ( − 1 ,   4 k ) The normal equation of is Y − 4 k = − 1 8 k ( X + 1 ) .    To make the normal cross the origin , The point of ( 0 ,   0 ) Satisfy the normal equation , have to k = ± 2 8    When x = 1 when , y ′ = − 8 k , So a little ( 1 ,   4 k ) The normal equation of is Y − 4 k = 1 8 k ( X − 1 ) .    To make the normal cross the origin , The point of ( 0 ,   0 ) Satisfy the normal equation , have to k = ± 2 8 , So when k = ± 2 8 when , The normal at the inflection point of the curve passes through the origin . \begin{aligned} &\ \ y'=2k(x^2-3)\cdot 2x=4kx(x^2-3),y''=4k(x^2-3)+4kx \cdot 2x=12kx(x-1)(x+1).\\\\ &\ \ Make y''=0, have to x_1=-1,x_2=1,y_1=4k,y_2=4k.\\\\ &\ \ When -\infty \lt x \lt -1 when ,y'' \gt 0, So the curve is (-\infty, \ -1] The top is concave ,\\\\ &\ \ When -1 \lt x \lt 1 when ,y'' \lt 0, So the curve is [-1,\ 1] It's convex ,\\\\ &\ \ When 1 \lt x \lt +\infty when ,y'' \gt 0, So the curve is [1, \ +\infty) Concave up , So point (-1, \ 4k) and (1, \ 4k) Is the inflection point of the curve .\\\\ &\ \ When x=-1 when ,y'=8k, So a little (-1, \ 4k) The normal equation of is Y-4k=-\frac{1}{8k}(X+1).\\\\ &\ \ To make the normal cross the origin , The point of (0, \ 0) Satisfy the normal equation , have to k=\pm \frac{\sqrt{2}}{8}\\\\ &\ \ When x=1 when ,y'=-8k, So a little (1, \ 4k) The normal equation of is Y-4k=\frac{1}{8k}(X-1).\\\\ &\ \ To make the normal cross the origin , The point of (0, \ 0) Satisfy the normal equation , have to k=\pm \frac{\sqrt{2}}{8}, So when k=\pm \frac{\sqrt{2}}{8} when , The normal at the inflection point of the curve passes through the origin . & \end{aligned} ​  y′=2k(x2−3)⋅2x=4kx(x2−3),y′′=4k(x2−3)+4kx⋅2x=12kx(x−1)(x+1).   Make y′′=0, have to x1​=−1,x2​=1,y1​=4k,y2​=4k.   When −∞<x<−1 when ,y′′>0, So the curve is (−∞, −1] The top is concave ,   When −1<x<1 when ,y′′<0, So the curve is [−1, 1] It's convex ,   When 1<x<+∞ when ,y′′>0, So the curve is [1, +∞) Concave up , So point (−1, 4k) and (1, 4k) Is the inflection point of the curve .   When x=−1 when ,y′=8k, So a little (−1, 4k) The normal equation of is Y−4k=−8k1​(X+1).   To make the normal cross the origin , The point of (0, 0) Satisfy the normal equation , have to k=±82​​   When x=1 when ,y′=−8k, So a little (1, 4k) The normal equation of is Y−4k=8k1​(X−1).   To make the normal cross the origin , The point of (0, 0) Satisfy the normal equation , have to k=±82​​, So when k=±82​​ when , The normal at the inflection point of the curve passes through the origin .​​

16.   set up y = f ( x ) stay x = x 0 Has a third-order continuous derivative in a neighborhood of , If f ′ ′ ( x 0 ) = 0 , and f ′ ′ ′ ( x 0 ) ≠ 0 ,        Try to ask ( x 0 ,   f ( x 0 ) ) Is it the inflection point ？ Why? ？ \begin{aligned}&16. \ set up y=f(x) stay x=x_0 Has a third-order continuous derivative in a neighborhood of , If f''(x_0)=0, and f'''(x_0) \neq 0,\\\\&\ \ \ \ \ \ Try to ask (x_0, \ f(x_0)) Is it the inflection point ？ Why? ？&\end{aligned} ​16.  set up y=f(x) stay x=x0​ Has a third-order continuous derivative in a neighborhood of , If f′′(x0​)=0, and f′′′(x0​)=0,       Try to ask (x0​, f(x0​)) Is it the inflection point ？ Why? ？​​

Explain ：

   It is known that f ′ ′ ′ ( x 0 ) ≠ 0 , hypothesis f ′ ′ ′ ( x 0 ) > 0 , because f ′ ′ ′ ( x 0 ) stay x = x 0 Continuous in a neighborhood of , So there must be δ > 0 ,    When x ∈ ( x 0 − δ ,   x 0 + δ ) when , f ′ ′ ′ ( x 0 ) > 0 , So in ( x 0 − δ ,   x 0 + δ ) Inside f ′ ′ ( x ) Monotonous increase ,    Again because f ′ ′ ( x 0 ) = 0 , So when x ∈ ( x 0 − δ ,   x 0 ) when , f ′ ′ ( x ) < f ′ ′ ( x 0 ) = 0 , be f ( x ) stay ( x 0 − δ ,   x 0 ) The figure inside is convex ,    When x ∈ ( x 0 ,   x 0 + δ ) when , f ′ ′ ( x ) > f ′ ′ ( x 0 ) = 0 , be f ( x ) stay ( x 0 ,   x 0 + δ ) The figure inside is concave , So point ( x 0 ,   f ( x 0 ) ) Is the inflection point of the curve . \begin{aligned} &\ \ It is known that f'''(x_0) \neq 0, hypothesis f'''(x_0) \gt 0, because f'''(x_0) stay x=x_0 Continuous in a neighborhood of , So there must be \delta \gt 0,\\\\ &\ \ When x \in (x_0-\delta, \ x_0+\delta) when ,f'''(x_0) \gt 0, So in (x_0-\delta, \ x_0+\delta) Inside f''(x) Monotonous increase ,\\\\ &\ \ Again because f''(x_0)=0, So when x \in (x_0-\delta, \ x_0) when ,f''(x) \lt f''(x_0)=0, be f(x) stay (x_0-\delta, \ x_0) The figure inside is convex ,\\\\ &\ \ When x \in (x_0, \ x_0+\delta) when ,f''(x) \gt f''(x_0)=0, be f(x) stay (x_0, \ x_0+\delta) The figure inside is concave , So point (x_0, \ f(x_0)) Is the inflection point of the curve . & \end{aligned} ​   It is known that f′′′(x0​)=0, hypothesis f′′′(x0​)>0, because f′′′(x0​) stay x=x0​ Continuous in a neighborhood of , So there must be δ>0,   When x∈(x0​−δ, x0​+δ) when ,f′′′(x0​)>0, So in (x0​−δ, x0​+δ) Inside f′′(x) Monotonous increase ,   Again because f′′(x0​)=0, So when x∈(x0​−δ, x0​) when ,f′′(x)<f′′(x0​)=0, be f(x) stay (x0​−δ, x0​) The figure inside is convex ,   When x∈(x0​, x0​+δ) when ,f′′(x)>f′′(x0​)=0, be f(x) stay (x0​, x0​+δ) The figure inside is concave , So point (x0​, f(x0​)) Is the inflection point of the curve .​​

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