Question

Let ​f(x)equalsx squared minus 6 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

Answer 1

To solve the problem, we will first define the functions you mentioned:

- \( f(x) = x^2 - 6x \)
- \( g(x) = 7 + x \)

Now, we can find the following:

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

The sum of the two functions is given by:

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

Calculating that:

\[
f(x) + g(x) = (x^2 - 6x) + (7 + x)
\]

Combining like terms:

\[
= x^2 - 6x + x + 7 = x^2 - 5x + 7
\]

Thus:

\[
(f + g)(x) = x^2 - 5x + 7
\]

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

The difference of the two functions is given by:

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

Calculating that:

\[
f(x) - g(x) = (x^2 - 6x) - (7 + x)
\]

Distributing the negative sign:

\[
= x^2 - 6x - 7 - x = x^2 - 7x - 7
\]

Thus:

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

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

The product of the two functions is given by:

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

Calculating that:

\[
f(x) \cdot g(x) = (x^2 - 6x)(7 + x)
\]

Using the distributive property (FOIL):

\[
= x^2 \cdot 7 + x^2 \cdot x - 6x \cdot 7 - 6x \cdot x
\]

Calculating each term:

\[
= 7x^2 + x^3 - 42x - 6x^2
\]

Combining like terms:

\[
= x^3 + (7x^2 - 6x^2) - 42x = x^3 + x^2 - 42x
\]

Thus:

\[
(f \cdot g)(x) = x^3 + x^2 - 42x
\]

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

The quotient of the two functions is given by:

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

Calculating it:

\[
\left( \frac{f}{g} \right)(x) = \frac{x^2 - 6x}{7 + x}
\]

Thus:

\[
\left( \frac{f}{g} \right)(x) = \frac{x^2 - 6x}{7 + x}
\]

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

The domain of a function is determined by the values of \( x \) for which the function is defined. For \( \left( \frac{f}{g} \right)(x) \), it is undefined when the denominator \( g(x) \) is equal to zero.

Setting the denominator to zero:

\[
7 + x = 0
\]

Solving for \( x \):

\[
x = -7
\]

Thus, the domain of \( \left( \frac{f}{g} \right) \) is all real numbers except \( -7 \), which can be written as:

\[
\text{Domain} = \{ x \in \mathbb{R} \mid x \neq -7 \}
\] or in interval notation:

\[
(-\infty, -7) \cup (-7, \infty)
\]

In summary:
- (a) \( (f + g)(x) = x^2 - 5x + 7 \)
- (b) \( (f - g)(x) = x^2 - 7x - 7 \)
- (c) \( (f \cdot g)(x) = x^3 + x^2 - 42x \)
- (d) \( \left( \frac{f}{g} \right)(x) = \frac{x^2 - 6x}{7 + x} \)
- (e) Domain: \( (-\infty, -7) \cup (-7, \infty) \)