p ( x ) = x 6 + 2 x 5 − 2 x 3 − x 2 p(x)=x^6+2x^5-2x^3-x^2 p ( x ) = x 6 + 2 x 5 − 2 x 3 − x 2 DomainL: ( − ∞ , ∞ ) (-\infin, \infin) ( − ∞ , ∞ )
As x → − ∞ , y → ∞ x\to-\infin,y\to\infin x → − ∞ , y → ∞
As x → ∞ , y → ∞ x\to\infin,y\to\infin x → ∞ , y → ∞
There are no asymptotes
p ( − x ) = ( − x ) 6 + 2 ( − x ) 5 − 2 ( − x ) 3 − ( − x ) 2 p(-x)=(-x)^6+2(-x)^5-2(-x)^3-(-x)^2 p ( − x ) = ( − x ) 6 + 2 ( − x ) 5 − 2 ( − x ) 3 − ( − x ) 2
= x 6 − 2 x 5 + 2 x 3 − x 2 =x^6-2x^5+2x^3-x^2 = x 6 − 2 x 5 + 2 x 3 − x 2 The function p ( x ) p(x) p ( x )
y y y x = 0 , y = p ( 0 ) = ( 0 ) 6 + 2 ( 0 ) 5 − 2 ( 0 ) 3 − ( 0 ) 2 = 0 x=0, y=p(0)=(0)^6+2(0)^5-2(0)^3-(0)^2=0 x = 0 , y = p ( 0 ) = ( 0 ) 6 + 2 ( 0 ) 5 − 2 ( 0 ) 3 − ( 0 ) 2 = 0
( 0 , 0 ) (0,0) ( 0 , 0 )
x x x y = 0 = > x 6 − 2 x 5 + 2 x 3 − x 2 = 0 y=0=>x^6-2x^5+2x^3-x^2=0 y = 0 => x 6 − 2 x 5 + 2 x 3 − x 2 = 0
x 2 ( x 4 − 2 x 3 + 2 x − 1 ) = 0 x^2(x^4-2x^3+2x-1)=0 x 2 ( x 4 − 2 x 3 + 2 x − 1 ) = 0
x 2 ( x + 1 ) 3 ( x − 1 ) = 0 x^2(x+1)^3(x-1)=0 x 2 ( x + 1 ) 3 ( x − 1 ) = 0
x 1 = x 2 = x 3 = − 1 , x 4 = x 5 = 0 , x 6 = 1 x_1= x_2=x_3=-1, x_4=x_5=0,x_6=1 x 1 = x 2 = x 3 = − 1 , x 4 = x 5 = 0 , x 6 = 1 ( − 1 , 0 ) , ( 0 , 0 ) , ( 1 , 0 ) (-1,0), (0,0), (1,0) ( − 1 , 0 ) , ( 0 , 0 ) , ( 1 , 0 )
The graph passes through the origin.
Find the first derivative
p ′ ( x ) = 6 x 5 + 10 x 4 − 6 x 2 − 2 x p'(x)=6x^5+10x^4-6x^2-2x p ′ ( x ) = 6 x 5 + 10 x 4 − 6 x 2 − 2 x Find the critical number(s)
p ′ ( x ) = 0 = > 6 x 5 + 10 x 4 − 6 x 2 − 2 x = 0 p'(x)=0=>6x^5+10x^4-6x^2-2x=0 p ′ ( x ) = 0 => 6 x 5 + 10 x 4 − 6 x 2 − 2 x = 0
2 x ( 3 x 4 + 5 x 3 − 3 x − 1 ) = 0 2x(3x^4+5x^3-3x-1)=0 2 x ( 3 x 4 + 5 x 3 − 3 x − 1 ) = 0
2 x ( x + 1 ) 2 ( 3 x 2 − x − 1 ) = 0 2x(x+1)^2(3x^2-x-1)=0 2 x ( x + 1 ) 2 ( 3 x 2 − x − 1 ) = 0
x 1 = x 2 = − 1 , x 3 = 0 , x_1=x_2=-1, x_3=0, x 1 = x 2 = − 1 , x 3 = 0 ,
x 4 = 1 − 13 6 , x 5 = 1 + 13 6 x_4=\dfrac{1-\sqrt{13}}{6}, x_5=\dfrac{1+\sqrt{13}}{6} x 4 = 6 1 − 13 , x 5 = 6 1 + 13 Critical numbers: − 1 , 1 − 13 6 , 0 , 1 + 13 6 -1,\dfrac{1-\sqrt{13}}{6}, 0,\dfrac{1+\sqrt{13}}{6} − 1 , 6 1 − 13 , 0 , 6 1 + 13
If x < − 1 , p ′ ( x ) < 0 , p ( x ) x<-1, p'(x)<0, p(x) x < − 1 , p ′ ( x ) < 0 , p ( x )
If − 1 < x < 1 − 13 6 , p ′ ( x ) < 0 , p ( x ) -1<x<\dfrac{1-\sqrt{13}}{6}, p'(x)<0, p(x) − 1 < x < 6 1 − 13 , p ′ ( x ) < 0 , p ( x )
If 1 − 13 6 < x < 0 , p ′ ( x ) > 0 , p ( x ) \dfrac{1-\sqrt{13}}{6}<x<0, p'(x)>0, p(x) 6 1 − 13 < x < 0 , p ′ ( x ) > 0 , p ( x )
If 0 < x < 1 + 13 6 , p ′ ( x ) < 0 , p ( x ) 0<x<\dfrac{1+\sqrt{13}}{6}, p'(x)<0, p(x) 0 < x < 6 1 + 13 , p ′ ( x ) < 0 , p ( x )
If x > 1 + 13 6 , p ′ ( x ) > 0 , p ( x ) x>\dfrac{1+\sqrt{13}}{6}, p'(x)>0, p(x) x > 6 1 + 13 , p ′ ( x ) > 0 , p ( x )
p ( − 1 ) = ( − 1 ) 6 + 2 ( − 1 ) 5 − 2 ( − 1 ) 3 − ( − 1 ) 2 = 0 p(-1)=(-1)^6+2(-1)^5-2(-1)^3-(-1)^2=0 p ( − 1 ) = ( − 1 ) 6 + 2 ( − 1 ) 5 − 2 ( − 1 ) 3 − ( − 1 ) 2 = 0
p ( 1 − 13 6 ) = ( 1 − 13 6 ) 6 + 2 ( 1 − 13 6 ) 5 p(\dfrac{1-\sqrt{13}}{6})=(\dfrac{1-\sqrt{13}}{6})^6+2(\dfrac{1-\sqrt{13}}{6})^5 p ( 6 1 − 13 ) = ( 6 1 − 13 ) 6 + 2 ( 6 1 − 13 ) 5
− 2 ( 1 − 13 6 ) 3 − ( 1 − 13 6 ) 2 -2(\dfrac{1-\sqrt{13}}{6})^3-(\dfrac{1-\sqrt{13}}{6})^2 − 2 ( 6 1 − 13 ) 3 − ( 6 1 − 13 ) 2
= − 587 − 143 13 1458 =-\dfrac{587-143\sqrt{13}}{1458} = − 1458 587 − 143 13
p ( 1 + 13 6 ) = ( 1 + 13 6 ) 6 + 2 ( 1 + 13 6 ) 5 p(\dfrac{1+\sqrt{13}}{6})=(\dfrac{1+\sqrt{13}}{6})^6+2(\dfrac{1+\sqrt{13}}{6})^5 p ( 6 1 + 13 ) = ( 6 1 + 13 ) 6 + 2 ( 6 1 + 13 ) 5
− 2 ( 1 + 13 6 ) 3 − ( 1 + 13 6 ) 2 = -2(\dfrac{1+\sqrt{13}}{6})^3-(\dfrac{1+\sqrt{13}}{6})^2= − 2 ( 6 1 + 13 ) 3 − ( 6 1 + 13 ) 2 =
= − 587 + 143 13 1458 =-\dfrac{587+143\sqrt{13}}{1458} = − 1458 587 + 143 13
p ( 0 ) = ( 0 ) 6 + 2 ( 0 ) 5 − 2 ( 0 ) 3 − ( 0 ) 2 = 0 p(0)=(0)^6+2(0)^5-2(0)^3-(0)^2=0 p ( 0 ) = ( 0 ) 6 + 2 ( 0 ) 5 − 2 ( 0 ) 3 − ( 0 ) 2 = 0 The function p ( x ) p(x) p ( x ) x = 0. x=0. x = 0.
The function p ( x ) p(x) p ( x ) − 587 − 143 13 1458 -\dfrac{587-143\sqrt{13}}{1458} − 1458 587 − 143 13 x = 1 − 13 6 . x=\dfrac{1-\sqrt{13}}{6}. x = 6 1 − 13 .
The function p ( x ) p(x) p ( x ) − 587 + 143 13 1458 -\dfrac{587+143\sqrt{13}}{1458} − 1458 587 + 143 13 x = 1 + 13 6 . x=\dfrac{1+\sqrt{13}}{6}. x = 6 1 + 13 .
Find the second derivative
p ′ ′ ( x ) = 30 x 4 + 40 x 3 − 12 x − 2 p''(x)=30x^4+40x^3-12x-2 p ′′ ( x ) = 30 x 4 + 40 x 3 − 12 x − 2 Find the point(s) of inflection.
p ′ ′ ( x ) = 0 = > 30 x 4 + 40 x 3 − 12 x − 2 = 0 p''(x)=0=>30x^4+40x^3-12x-2=0 p ′′ ( x ) = 0 => 30 x 4 + 40 x 3 − 12 x − 2 = 0
2 ( x + 1 ) ( 15 x 3 + 5 x 2 − 5 x − 1 ) = 0 2(x+1)(15x^3+5x^2-5x-1)=0 2 ( x + 1 ) ( 15 x 3 + 5 x 2 − 5 x − 1 ) = 0
x 1 = − 1 , x 2 ≈ − 0.6795 , x 3 ≈ − 0.1848 , x_1=-1, x_2\approx-0.6795, x_3\approx-0.1848, x 1 = − 1 , x 2 ≈ − 0.6795 , x 3 ≈ − 0.1848 ,
x 4 ≈ 0.5310 x_4\approx0.5310 x 4 ≈ 0.5310 The function p ( x ) p(x) p ( x ) x = − 1 , x = − 0.6795 , x=-1, x=-0.6795, x = − 1 , x = − 0.6795 ,
x = − 0.1848 , 0.5310. x=-0.1848,0.5310. x = − 0.1848 , 0.5310.
If x < − 1 , p ′ ′ ( x ) > 0 , p ( x ) x<-1, p''(x)>0, p(x) x < − 1 , p ′′ ( x ) > 0 , p ( x )
If − 1 < x < − 0.6795 , p ′ ′ ( x ) < 0 , p ( x ) -1<x<-0.6795, p''(x)<0, p(x) − 1 < x < − 0.6795 , p ′′ ( x ) < 0 , p ( x )
If − 0.6795 < x < − 0.1848 , p ′ ′ ( x ) > 0 , p ( x ) -0.6795<x<-0.1848, p''(x)>0, p(x) − 0.6795 < x < − 0.1848 , p ′′ ( x ) > 0 , p ( x )
If − 0.1848 < x < 0.5310 , p ′ ′ ( x ) < 0 , p ( x ) -0.1848<x<0.5310, p''(x)<0, p(x) − 0.1848 < x < 0.5310 , p ′′ ( x ) < 0 , p ( x )
If x > 0.5310 , p ′ ′ ( x ) > 0 , p ( x ) x>0.5310, p''(x)>0, p(x) x > 0.5310 , p ′′ ( x ) > 0 , p ( x )
Graph of the function
#340153 on Dec 2023
Comments