Chapter 1.4, Problem 93E

To determine

Why are the domains of $f$ and $g$ different?.

Answer to Problem 93E

Both the functions has different domain.

Explanation of Solution

Given information:

Consider

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

Calculation:

Consider the given functions.

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

Now check if both the function has different domains.

Now find the domain of,

  $f\left(x\right)=\sqrt{x-1}$

The radical should be non-negative thus,

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

The domain of this function is all real values of $x$ for $x\ge 1$

Now find the domain of,

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

We will include those values where denominator is non-zero and non negative.

  $\begin{array}{l}g\left(x\right)=\frac{1}{\sqrt{x-1}}\\ \sqrt{x-1}\ge 0\end{array}$

Square both sides,

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

The denominator is of square root form so $x$ has to be non-negative

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

So, $x=1$ the denominator become zero , then we will exclude $x=1$ .

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

Hence, both the functions has different domain.