Answer to Question #286993 in Calculus for Jishah

Question

graphing p(x)=x⁶+2x⁵-2x³-x²

Answer 1

$$p(x)=x^6+2x^5-2x^3-x^2$$

DomainL: $(-\infin, \infin)$

As $x\to-\infin,y\to\infin$

As $x\to\infin,y\to\infin$

There are no asymptotes

$$p(-x)=(-x)^6+2(-x)^5-2(-x)^3-(-x)^2$$

$$=x^6-2x^5+2x^3-x^2$$

The function $p(x)$ is neither even nor odd.

$y$ -intercept: $x=0, y=p(0)=(0)^6+2(0)^5-2(0)^3-(0)^2=0$

$(0,0)$

$x$ -intercept(s): $y=0=>x^6-2x^5+2x^3-x^2=0$

$$x^2(x^4-2x^3+2x-1)=0$$

$$x^2(x+1)^3(x-1)=0$$

$$x_1= x_2=x_3=-1, x_4=x_5=0,x_6=1$$

$(-1,0), (0,0), (1,0)$

The graph passes through the origin.

Find the first derivative

$$p'(x)=6x^5+10x^4-6x^2-2x$$

Find the critical number(s)

$$p'(x)=0=>6x^5+10x^4-6x^2-2x=0$$

$$2x(3x^4+5x^3-3x-1)=0$$

$$2x(x+1)^2(3x^2-x-1)=0$$

$$x_1=x_2=-1, x_3=0,$$

$$x_4=\dfrac{1-\sqrt{13}}{6}, x_5=\dfrac{1+\sqrt{13}}{6}$$

Critical numbers: $-1,\dfrac{1-\sqrt{13}}{6}, 0,\dfrac{1+\sqrt{13}}{6}$

If $x<-1, p'(x)<0, p(x)$ decreases.

If $-1<x<\dfrac{1-\sqrt{13}}{6}, p'(x)<0, p(x)$ decreases.

If $\dfrac{1-\sqrt{13}}{6}<x<0, p'(x)>0, p(x)$ increases.

If $0<x<\dfrac{1+\sqrt{13}}{6}, p'(x)<0, p(x)$ decreases.

If $x>\dfrac{1+\sqrt{13}}{6}, p'(x)>0, p(x)$ increases.

$$p(-1)=(-1)^6+2(-1)^5-2(-1)^3-(-1)^2=0$$

$$p(\dfrac{1-\sqrt{13}}{6})=(\dfrac{1-\sqrt{13}}{6})^6+2(\dfrac{1-\sqrt{13}}{6})^5$$

$$-2(\dfrac{1-\sqrt{13}}{6})^3-(\dfrac{1-\sqrt{13}}{6})^2$$

$$=-\dfrac{587-143\sqrt{13}}{1458}$$

$$p(\dfrac{1+\sqrt{13}}{6})=(\dfrac{1+\sqrt{13}}{6})^6+2(\dfrac{1+\sqrt{13}}{6})^5$$

$$-2(\dfrac{1+\sqrt{13}}{6})^3-(\dfrac{1+\sqrt{13}}{6})^2=$$

$$=-\dfrac{587+143\sqrt{13}}{1458}$$

$$p(0)=(0)^6+2(0)^5-2(0)^3-(0)^2=0$$

The function $p(x)$ has a local maximum with value of at $x=0.$

The function $p(x)$ has a local minimum with value of $-\dfrac{587-143\sqrt{13}}{1458}$ at $x=\dfrac{1-\sqrt{13}}{6}.$

The function $p(x)$ has a local minimum with value of $-\dfrac{587+143\sqrt{13}}{1458}$ at $x=\dfrac{1+\sqrt{13}}{6}.$

Find the second derivative

$$p''(x)=30x^4+40x^3-12x-2$$

Find the point(s) of inflection.

$$p''(x)=0=>30x^4+40x^3-12x-2=0$$

$$2(x+1)(15x^3+5x^2-5x-1)=0$$

$$x_1=-1, x_2\approx-0.6795, x_3\approx-0.1848,$$

$$x_4\approx0.5310$$

The function $p(x)$ has the inflection points at $x=-1, x=-0.6795,$

$x=-0.1848,0.5310.$

If $x<-1, p''(x)>0, p(x)$ is concave up.

If $-1<x<-0.6795, p''(x)<0, p(x)$ is concave down.

If $-0.6795<x<-0.1848, p''(x)>0, p(x)$ is concave up.

If $-0.1848<x<0.5310, p''(x)<0, p(x)$ is concave down.

If $x>0.5310, p''(x)>0, p(x)$ is concave up.

Graph of the function