Question

level 4: open ended questions

1. sketch the graph of a single function with a domain $(-\infty,\infty)$ that has all of the following characteristics

a. a cubic function
b. inflection point at $(2,5)$
c decreasing and concave up on the interval $(-\infty,2)$
d. decreasing and concave up on the interval $(2,\infty)$

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 \(x>2\))