Chapter 2.4, Problem 41E

To determine

To find: $\left[f\cdot g\right]\left(x\right)$ and $\left[\frac{f}{g}\right]\left(x\right)$

Answer to Problem 41E

  $\begin{array}{c}\left[f\cdot g\right]\left(x\right)=x^5-3x^4+7x^3\\ \left[\frac{f}{g}\right]\left(x\right)=\frac{x^3}{x^2-3x+7}\end{array}$

Explanation of Solution

Given information:

The given functions are as follows:

  $\begin{array}{l}f\left(x\right)=x^3\\ g\left(x\right)=x^2-3x+7\end{array}$

Calculation:

Solve the $\left[f\cdot g\right]\left(x\right)$ as follows:

  $\begin{array}{c}\left[f\cdot g\right]\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(x^3\right)\left(x^2-3x+7\right)\\ =x^5-3x^4+7x^3\end{array}$

Thus, the value of $\left[f\cdot g\right]\left(x\right)$ is $x^5-3x^4+7x^3$ .

Solve the $\left[\frac{f}{g}\right]\left(x\right)$ as follows:

  $\begin{array}{c}\left[\frac{f}{g}\right]\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{x^3}{x^2-3x+7}\end{array}$

Thus, the value of $\left[\frac{f}{g}\right]\left(x\right)$ is $\frac{x^3}{x^2-3x+7}$ .