Chapter 2.2, Problem 79E

To determine

To graph: The polynomial function that satisfy the given conditions. If not possible then to explain the reason.

Explanation of Solution

Given data:

The polynomial function $f\left(x\right)$ is a third-degree polynomial with two real zeros and a negative leading coefficient.

Concept used:

If $x={\alpha }_1,x={\alpha }_2,x={\alpha }_3,\dots x={\alpha }_n,$ are the zeros of a polynomial function $f\left(x\right)$ (say) of multiplicities $k_1,k_2,k_3,\dots ,k_n$ then ${\left(x-{\alpha }_1\right)}^{k_1}$ , ${\left(x-{\alpha }_2\right)}^{k_2}$ , ${\left(x-{\alpha }_3\right)}^{k_3}$$\dots ,{\left(x-{\alpha }_n\right)}^{k_n}$ are the factors of $f\left(x\right)$ that is a polynomial function of degree n given by $f\left(x\right)={\left(x-{\alpha }_1\right)}^{k_1}{\left(x-{\alpha }_2\right)}^{k_2}{\left(x-{\alpha }_3\right)}^{k_3}\dots {\left(x-{\alpha }_n\right)}^{k_n}$ .

Calculations:

Let $f\left(x\right)$ be the polynomial function having zeros $x={\alpha }_1$ of multiplicity $k_1$ and $x={\alpha }_2$ of multiplicity $k_2$ with leading coefficient negative and of degree $\left(k_1+k_2\right)$ given by

  $f\left(x\right)={\left(ax-{\alpha }_1\right)}^{k_1}{\left(x-{\alpha }_2\right)}^{k_2}$ ...............(1)

Let we choose arbitrarily two zeros of $f\left(x\right)$ , ${\alpha }_1=-3$ zeros of multiplicity $k_1=1$ and ${\alpha }_2=1$ of multiplicity $k_2=2$ ,then

  $f\left(x\right)={\left(ax-(-3)\right)}^1{\left(x-2\right)}^2$

  $\Rightarrow f\left(x\right)=\left(ax+3\right)\left(x^2-4x+4\right)$

The leading coefficient of function must be negative so we take $a=-1$

  $\Rightarrow f\left(x\right)=\left(-x+3\right)\left(x^2-4x+4\right)$

  $\Rightarrow f\left(x\right)=-x^3+x^2+8x-12$ .

The graph of the required function $f\left(x\right)=-x^3+x^2+8x-12$ is shown in Fugure-1 here.

  

Interpretation of the Graph:

The function $f\left(x\right)$ intersects x -axis at point $\left(-3,0\right)$ and touches at point $\left(2,0\right)$ , so the point $x=2$ is multiple root (double root). Since, function $f\left(x\right)$ is of odd degree and leading coefficient is negative so graph of $f\left(x\right)$ rises to the left and falls to the right.