Chapter 2.2, Problem 81E

To determine

To graph: The polynomial function $f\left(x\right)$ 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 fifth-degree with three real zeros and a positive 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 $k_1$ , $x={\alpha }_2$ of multiplicity $k_2$ and $x={\alpha }_3$ of multiplicity $k_3$ with leading coefficient positivegiven by

  $f\left(x\right)={\left(x-{\alpha }_1\right)}^{k_1}{\left(x-{\alpha }_2\right)}^{k_2}{\left(x-{\alpha }_3\right)}^{k_3}$....................(1)

Let we choose arbitrarily three zeros of $f\left(x\right)$ , ${\alpha }_1=-1$ zeros of multiplicity $k_1=2$ and ${\alpha }_2=-3$ of multiplicity $k_2=1$ ,and ${\alpha }_3=2$ of multiplicity $k_2=2$ , then

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

  $\Rightarrow f\left(x\right)=\left(x^2+2x+1\right)\left(x+3\right)\left(x^2-4x+4\right)$

  $\Rightarrow f\left(x\right)=x^5+x^4-9x^3-5x^2+16x+12$ .

The graph of the requiredfunction $f\left(x\right)=x^5+x^4-9x^3-5x^2+16x+12$ is shown in Fugure-1.

  

Interpretation of the Graph:

The function $f\left(x\right)$ intersects x -axis at point $\left(-3,0\right)$ and touches at points $\left(-1,0\right)$ and $\left(2,0\right)$ , so points $x=-1$ and $x=2$ are multiple root (double root). Since, function $f\left(x\right)$ is of odd degree and leading coefficient is positive so graph of $f\left(x\right)$ falls to the left and rises to the right.