level 4: open ended questions\n2. sketch the graph of a single function with a domain $(-\\infty,\\infty)$…

level 4: open ended questions\n2. sketch the graph of a single function with a domain $(-\\infty,\\infty)$ that has all of the following characteristics\n a. a cubic function\n b. inflection point at $(2,5)$\n c decreasing and concave up on the interval $(-\\infty,2)$\n d. decreasing and concave up on the interval $(2,\\infty)$

level 4: open ended questions\n2. sketch the graph of a single function with a domain $(-\\infty,\\infty)$ that has all of the following characteristics\n a. a cubic function\n b. inflection point at $(2,5)$\n c decreasing and concave up on the interval $(-\\infty,2)$\n d. decreasing and concave up on the interval $(2,\\infty)$

Answer

Explanation:

Step1: Recall properties of cubic functions

A general cubic function is (y = ax^{3}+bx^{2}+cx + d). The second - derivative (y''=6ax + 2b). The inflection point occurs where (y'' = 0). Given the inflection point at (x = 2), we set (6a\times2+2b=0), so (b=-6a). Let (a = 1), then (b=-6).

Step2: Consider the first - derivative for monotonicity

The first - derivative (y'=3ax^{2}+2bx + c). Since the function is decreasing on ((-\infty,\infty)), (y'\leq0) for all (x). For (a = 1) and (b=-6), (y'=3x^{2}-12x + c). To have (y'\leq0) for all (x), the discriminant (\Delta=( - 12)^{2}-12c\leq0). Let (c = 12), then (y'=3x^{2}-12x + 12=3(x - 2)^{2}).

Step3: Use the inflection - point to find the function

We know the inflection point ((2,5)). Let (y=(x - 2)^{3}+k). Substitute (x = 2,y = 5) into the function, we get (k = 5). So (y=(x - 2)^{3}+5=x^{3}-6x^{2}+12x - 8 + 5=x^{3}-6x^{2}+12x - 3)

Step4: Sketch the graph

  • For (x<2): The second - derivative (y''=6x-12). When (x<2), (y''<0) (concave down was a wrong initial thought, correct: since (y=(x - 2)^{3}+5), (y'=3(x - 2)^{2}\geq0) (error in previous step, correct: (y' = 3(x - 2)^{2}), and the function (y=(x - 2)^{3}+5) has (y'=3(x - 2)^{2}). The function is decreasing when (y'<0) (correction: (y=(x - 2)^{3}+5), (y'=3(x - 2)^{2}\geq0) is wrong. Correct (y=-(x - 2)^{3}+5), (y'=-3(x - 2)^{2}\leq0) for all (x), (y''=-6(x - 2)). When (x<2), (y''>0) (concave up), when (x>2), (y''<0) (concave down). Wait, no: if (y =-(x - 2)^{3}+5=-x^{3}+6x^{2}-12x + 8 + 5=-x^{3}+6x^{2}-12x+13), (y'=-3x^{2}+12x - 12=-3(x - 2)^{2}\leq0) (function is decreasing for all (x)), (y''=-6x + 12). When (x<2), (y''>0) (concave up), when (x>2), (y''<0) (concave down). Plot the inflection point ((2,5)). As (x\to-\infty), (y\to+\infty) (since (y=-x^{3}+6x^{2}-12x + 13), the leading term (-x^{3}) dominates), as (x\to+\infty), (y\to-\infty). Mark the inflection point ((2,5)). Since (y'=-3(x - 2)^{2}), the slope is non - positive everywhere.

Answer:

Sketch the graph of (y=-(x - 2)^{3}+5=-x^{3}+6x^{2}-12x + 13). Mark the point ((2,5)). The function is decreasing (since (y'=-3(x - 2)^{2}\leq0) for all (x)). For (x<2), the graph is concave up ((y''=-6x + 12>0) when (x<2)) and for (x>2), the graph is concave down ((y''=-6x + 12<0) when (x>2))