Chapter 1.4, Problem 89E

To determine

The error of the function $f\left(x\right)=\sqrt{x-1}andg\left(x\right)=\frac{1}{\sqrt{x-1}}$

Answer to Problem 89E

The domain of this function is all real values of x where $x>1$

Explanation of Solution

Given:

The error of the function $f\left(x\right)=\sqrt{x-1}andg\left(x\right)=\frac{1}{\sqrt{x-1}}$

Formula used:

The domain of the function is calculated by using the radical that should be non- negative.

Calculation:

Consider the following functions

  $f\left(x\right)=\sqrt{x-1}andg\left(x\right)=\frac{1}{\sqrt{x-1}}$

To check whether both the function has different domains.

Finding the domain of following function as,

  $f\left(x\right)=\sqrt{x-1}$

Now the radical should be non-negative, thus

  $\sqrt{x-1}\ge 0$

Squaring both sides

  $\begin{array}{l}x-1\ge 0\\ x\ge 1\\ \text{For}x=1\\ f\left(1\right)=\sqrt{1-1}\\ f\left(1\right)=0\end{array}$

Which holds, Hence, the domain of this function is all real values of x where $x\ge 1$

Now finding the domain of following function as,

  $g\left(x\right)=\frac{1}{\sqrt{x-1}}$

It includes only those values in the domain where denominator is non-zero and non- negative.

  $\sqrt{x-1}\ge 0$

Squaring both sides

  $\begin{array}{l}x-1\ge 0\\ x\ge 1\end{array}$

As the denominator is of square root form, so x has to be non-negative.

  $\begin{array}{l}x-1=0\\ x=1\end{array}$

For $x=1,$ the denominator becomes zero.

So, exclude $x=1$

Hence, the domain of this function is all real values of x where $x>1$

Thus, theabove conclude that both the functions have different domains.

Conclusion:

The domain of this function is all real values of x where $x>1$ .Thus, it concludes that both the functions have different domains.