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