Question

Let ​f(x)equalsx squared minus 3 x and ​g(x)equals7 plus x. Find the following.
​(a) left parenthesis f plus g right parenthesis left parenthesis x right parenthesis
​(b) left parenthesis f minus g right parenthesis left parenthesis x right parenthesis
​(c) left parenthesis f times g right parenthesis left parenthesis x right parenthesis
​(d) left parenthesis StartFraction f Over g EndFraction right parenthesis
​(x)
​(e) The domain of StartFraction f Over g EndFraction
Question content area bottom
Part 1
​(a) left parenthesis f plus g right parenthesis left parenthesis x right parenthesisequals
  
enter your response here ​(Simplify your answer. Do not​ factor.)

Answer 1

To solve the problem, we need to find various operations involving the functions \( f(x) \) and \( g(x) \).

Given:

- \( f(x) = x^2 - 3x \)
- \( g(x) = 7 + x \)

Let's compute each part:

### (a) \( (f + g)(x) \)

To find \( (f + g)(x) \), we add the two functions:

\[
(f + g)(x) = f(x) + g(x)
\]

Substituting in the expressions for \( f(x) \) and \( g(x) \):

\[
(f + g)(x) = (x^2 - 3x) + (7 + x)
\]

Now, combine like terms:

\[
= x^2 - 3x + 7 + x
\]
\[
= x^2 - 2x + 7
\]

Thus,

\[
\boxed{x^2 - 2x + 7}
\]

### (b) \( (f - g)(x) \)

To find \( (f - g)(x) \), we subtract \( g(x) \) from \( f(x) \):

\[
(f - g)(x) = f(x) - g(x)
\]

Substituting the expressions:

\[
(f - g)(x) = (x^2 - 3x) - (7 + x)
\]

Distributing the negative sign:

\[
= x^2 - 3x - 7 - x
\]

Combining like terms:

\[
= x^2 - 4x - 7
\]

Thus,

\[
\boxed{x^2 - 4x - 7}
\]

### (c) \( (f \times g)(x) \)

To find \( (f \times g)(x) \), we multiply the two functions:

\[
(f \times g)(x) = f(x) \cdot g(x)
\]

Substituting the expressions:

\[
(f \times g)(x) = (x^2 - 3x)(7 + x)
\]

Using the distributive property:

\[
= x^2 \cdot (7 + x) - 3x \cdot (7 + x)
\]
\[
= 7x^2 + x^3 - 21x - 3x^2
\]
\[
= x^3 + 4x^2 - 21x
\]

Thus,

\[
\boxed{x^3 + 4x^2 - 21x}
\]

### (d) \( \left(\frac{f}{g}\right)(x) \)

To find \( \left(\frac{f}{g}\right)(x) \), we divide \( f(x) \) by \( g(x) \):

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

Substituting the expressions:

\[
\left(\frac{f}{g}\right)(x) = \frac{x^2 - 3x}{7 + x}
\]

Thus,

\[
\boxed{\frac{x^2 - 3x}{7 + x}}
\]

### (e) The domain of \( \left(\frac{f}{g}\right) \)

The domain of \( \left(\frac{f}{g}\right)(x) \) is determined by the values of \( x \) for which \( g(x) \) is not equal to zero:

\[
g(x) = 7 + x \neq 0 \implies x \neq -7
\]

Therefore, the domain of \( \left(\frac{f}{g}\right)(x) \) in interval notation is:

\[
\boxed{(-\infty, -7) \cup (-7, \infty)}
\]