level 4: open ended questions\n4. sketch the graph of a single function with a domain $(-\\infty,\\infty)$…

Explanation:

Step1: Determine the general form of the cubic function

Since the zeros are at (x = - 1), (x = 0), and (x = 1), the cubic function can be written in factored form as (y=a(x + 1)x(x - 1)=a(x^{3}-x)), where (a) is a non - zero constant.

Step2: Use the end - behavior to find the value of (a)

The end - behavior is (x\rightarrow\infty,f(x)\rightarrow-\infty) and (x\rightarrow-\infty,f(x)\rightarrow\infty). For a cubic function (y = ax^{3}+bx^{2}+cx + d), the leading term (ax^{3}) determines the end - behavior. When (x\rightarrow\infty), if (y\rightarrow-\infty) and (x\rightarrow-\infty), (y\rightarrow\infty), then (a<0). Let (a=-1), so the function is (y=-x^{3}+x).

Step3: Check the inflection point and concavity

First, find the second derivative. The first derivative (y^\prime=-3x^{2}+1), and the second derivative (y^{\prime\prime}=-6x). Set (y^{\prime\prime}=0), then (-6x = 0) gives (x = 0). When (x = 0), (y=0), so the inflection point is ((0,0)). For the concavity:

• When (x\in(-\infty,0)), (y^{\prime\prime}=-6x>0), so the function is concave up on ((-\infty,0)).
• When (x\in(0,\infty)), (y^{\prime\prime}=-6x<0), so the function is concave down on ((0,\infty)).

Step4: Sketch the graph

• Plot the zeros at (x=-1), (x = 0), and (x = 1).
• Since the function is (y=-x^{3}+x), when (x=-2), (y=-(-2)^{3}+(-2)=8 - 2=6); when (x = 2), (y=-2^{3}+2=-8 + 2=-6).
• Mark the inflection point ((0,0)).
• Use the end - behavior ((x\rightarrow\infty,y\rightarrow-\infty) and (x\rightarrow-\infty,y\rightarrow\infty)) and the concavity information to draw a smooth curve passing through the plotted points.

Answer:

Sketch the graph of the function (y=-x^{3}+x) with zeros at (x=-1), (x = 0), (x = 1), inflection point at ((0,0)), concave up on ((-\infty,0)), concave down on ((0,\infty)) and the given end - behavior.