Chapter 5.4, Problem 28WE

a.

To determine

To find the domain of the given rational function.

Answer to Problem 28WE

Domain: $(-\infty ,1)\cup (1,\infty )$

Explanation of Solution

Given:

Rational function: $h(t)=\frac{t^3+4t^2-t-4}{t^3-t^2+t-1}$

Concept used:

To find the domain of the rational function, the following steps are to be followed:

1. Equate the denominator equal to zero.
2. Solve and find the t values that make the denominator equal to zero.
3. The domain is all the real numbers except for those t values equal to zero.

Calculation:

The given rational function is $h(t)=\frac{t^3+4t^2-t-4}{t^3-t^2+t-1}$

The denominator of the given function is $t^3-t^2+t-1.$

Equating the denominator to zero,

  $\Rightarrow t^3-t^2+t-1=0$

Grouping the terms,

  $\begin{array}{l}\Rightarrow (t^3-t^2)+(t-1)=0\\ \Rightarrow t^2(t-1)+(t-1)=0\\ \Rightarrow (t^2+1)(t-1)=0\\ \Rightarrow (t^2+1)\text{and}(t-1)=0\end{array}$

When $(t^2+1)=0,$

Subtracting 1 on both sides,

  $\begin{array}{l}\Rightarrow t^2+1-1=0-1\\ \Rightarrow t^2=-1\end{array}$

Taking square root on both sides,

  $\begin{array}{l}\Rightarrow \sqrt{t^2}=\sqrt{-1}\\ \Rightarrow t=\pm i\end{array}$

When $(t-1)=0,$

Adding 1 on both sides,

  $\begin{array}{l}\Rightarrow t-1+1=0+1\\ \Rightarrow t=1\end{array}$

Imaginary roots cannot be plotted on real plane.

Hence, the domain of the function is all real numbers except $t=1$ .

Conclusion:

Therefore, the domain of the function is a $(-\infty ,1)\cup (1,\infty ).$

b.

To determine

To find the zeros of the function if any.

Answer to Problem 28WE

  $\begin{array}{l}t=-4\\ t=\pm 1\end{array}$

Explanation of Solution

Given:

Rational function: $h(t)=\frac{t^3+4t^2-t-4}{t^3-t^2+t-1}$

Concept used:

To find the zeros of the rational function,

1. Equate $h(t)=0.$
2. Simplify and find the values of t which are the zeros of the rational function.

Calculation:

Equate the rational function $h(t)$ to zero.

  $\begin{array}{l}\Rightarrow h(t)=0\\ \Rightarrow \frac{t^3+4t^2-t-4}{t^3-t^2+t-1}=0\end{array}$

Cross multiplying,

  $\begin{array}{l}\Rightarrow t^3+4t^2-t-4=(t^3-t^2+t-1)(0)\\ \Rightarrow t^3+4t^2-t-4=0\end{array}$

Grouping the terms,

  $\begin{array}{l}\Rightarrow t^3+4t^2-t-4=0\\ \Rightarrow t^2(t+4)-(t+4)=0\\ \Rightarrow (t^2-1)(t+4)=0\\ \Rightarrow (t^2-1)\text{and}(t+4)=0\end{array}$

When $(t^2-1)=0,$

Adding 1 on both sides,

  $\begin{array}{l}\Rightarrow t^2-1+1=0+1\\ \Rightarrow t^2=1\end{array}$

Taking square root on both sides,

  $\begin{array}{l}\Rightarrow \sqrt{t^2}=\sqrt{1}\\ \Rightarrow t=\pm 1\end{array}$

When $(t+4)=0,$

Subtracting4 on both sides,

  $\begin{array}{l}\Rightarrow t+4-4=0-4\\ \Rightarrow t=-4\end{array}$

Conclusion:

Therefore, the zeroes of the rational function are $t=-4\text{ and }t=\pm 1$ .