Chapter 4, Problem 23CR

Suppose that $f\left(x\right)=x^2-5x+1$ and $g\left(x\right)=-4x-7$ .

Find $f+g$ and state its domain.

Find $\frac{f}{g}$ and its domain.

To determine

The function $f+g$ and find its domain, where $f\left(x\right)=x^2-5x+1$ and $g\left(x\right)=-4x-7$

Answer to Problem 23CR

Solution:

$f+g=x^2-9x-6$

The domain of $f+g$ is a set of all real numbers.

Explanation of Solution

Given information:

The function $f\left(x\right)=x^2-5x+1$ and $g\left(x\right)=-4x-7$

Explanation:

$f+g=f\left(x\right)+g\left(x\right)$

By substituting $f\left(x\right)$ and $g\left(x\right)$ in $f\left(x\right)+g\left(x\right)$

$f+g=\left(x^2-5x+1\right)+\left(-4x-7\right)$

$\Rightarrow f+g=x^2-5x-4x+1-7$

$\Rightarrow f+g=x^2-9x-6$

Thus, $f+g=x^2-9x-6$

Domain of $f\left(x\right)+g\left(x\right)$ is the intersection of the domain of $f\left(x\right)$ and domain of $g\left(x\right)$.

Since $f\left(x\right)$ and $g\left(x\right)$ are both polynomials, their domains are all real numbers.

Thus, domain of $f+g$ is set of all real numbers.

To determine

The function $\frac{f}{g}$ and find its domain, where $f\left(x\right)=x^2-5x+1$ and $g\left(x\right)=-4x-7$

Answer to Problem 23CR

Solution:

$\frac{f}{g}=\frac{x^2-5x+1}{-4x-7}$

The domain of the function $\frac{f}{g}$ is $\left(-\infty ,-\frac{7}{4}\right)\cup \left(-\frac{7}{4},\infty \right)$

Explanation of Solution

Given information:

The function $f\left(x\right)=x^2-5x+1$ and $g\left(x\right)=-4x-7$

Explanation:

$\frac{f}{g}=\frac{f\left(x\right)}{g\left(x\right)}$

By substituting $f\left(x\right)$ and $g\left(x\right)$ in $\frac{f}{g}$

$\frac{f}{g}=\frac{x^2-5x+1}{-4x-7}$

Thus, $\frac{f}{g}=\frac{x^2-5x+1}{-4x-7}$

Now, to find domain of $\frac{f}{g}=\frac{x^2-5x+1}{-4x-7}$

Domain of $\frac{f\left(x\right)}{g\left(x\right)}$ is intersection of the domain of $f\left(x\right)$ and domain of $g\left(x\right)$ along with confirmation of non-zero $g\left(x\right)$

Here, $f\left(x\right)$ and $g\left(x\right)$ are both polynomials, their domains are all real numbers.

Thus, $\frac{f}{g}$ is defined for all values except when $-4x-7=0$

By solving $-4x-7=0$, $x=-\frac{7}{4}$.

Hence, the function $\frac{f}{g}$ is defined for all real numbers except for $x=-\frac{7}{4}$

Hence, the domain of the function $\frac{f}{g}$ is $\left(-\infty ,-\frac{7}{4}\right)\cup \left(-\frac{7}{4},\infty \right)$