Chapter 1.2, Problem 93E

To determine

Todescribe:The error in finding the common domain of two functions.

Answer to Problem 93E

The error is that member $x=1$ is included in the domain set of function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ for which $g\left(x\right)$ is undefined.

Explanation of Solution

Given information:

Claim: The functions $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ have same domain, which is the set of all real numbers x such that $x\ge 1$ .

Conceptused:

The domain of a function is the set on which the function is defined.

The function $f\left(x\right)=\sqrt{x-1}$ is defined for all real $x\ge 1$ as for $x<1$ , values of $f\left(x\right)=\sqrt{x-1}$ are imaginary.

The function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ is defined for all real $x>1$ as for $x<1$ , values of $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ are imaginary. However, for $x=1$ , the function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ is undefined.

Thus, the domain of function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ is all real $x>1$ instead of all real numbers $x\ge 1$ that shows the domains of functions $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ are not same.

The error is that member $x=1$ is included in the domain set of function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ for which $g\left(x\right)$ is undefined.

Conclusion:

The error is that member $x=1$ is included in the domain set of function $g\left(x\right)=\frac{1}{\sqrt{x-1}}$ for which $g\left(x\right)$ is undefined.