Chapter 3, Problem 10CLT

a .

To determine

To find: $\left(f+g\right)\left(x\right)$ where $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=x^2+1$

Answer to Problem 10CLT

  $\left(f+g\right)\left(x\right)=\sqrt{x-1}+x^2+1$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\sqrt{x-1}+x^2+1\end{array}$

b.

To determine

To find: $\left(f-g\right)\left(x\right)$ where $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=x^2+1$

Answer to Problem 10CLT

  $\left(f-g\right)\left(x\right)=\sqrt{x-1}-x^2-1$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\sqrt{x-1}-\left(x^2+1\right)\\ =\sqrt{x-1}-x^2-1\end{array}$

c .

To determine

To find: $\left(fg\right)\left(x\right)$ where $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=x^2+1$

Answer to Problem 10CLT

  $\left(fg\right)\left(x\right)=x^2\sqrt{x-1}+\sqrt{x-1}$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(\sqrt{x-1}\right)\cdot \left(x^2+1\right)\\ =x^2\sqrt{x-1}+\sqrt{x-1}\end{array}$

d.

To determine

To find: $\left(\frac{f}{g}\right)\left(x\right)$ where $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=x^2+1$

Answer to Problem 10CLT

  $\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x-1}}{x^2+1}$

Domain: All real numbers x such that $x\ge 1$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\sqrt{x-1}}{x^2+1}\end{array}$

Now, the function $f\left(x\right)=\sqrt{x-1}$ has domain all real numbers x such that $x\ge 1$ .

The domain of $g\left(x\right)$ is the set of real numbers.

But the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is the domain common to both $f\left(x\right)$ and $g\left(x\right)$ but with condition $g\left(x\right)\ne 0$ .

Now, here denominator is never zero.

So, the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is same as domain of $f\left(x\right)$ .

Hence, the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is all real numbers x such that $x\ge 1$ .