Question

Let ​f(x)equalsx squared minus 6 x and ​g(x)equals7 plus x. Find the following.
​(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 problems step-by-step, we have the functions defined as:

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

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

This means we need to multiply \(f(x)\) and \(g(x)\):

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

Now, we can expand this product:

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

Thus,

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

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

This means we need to divide \(f(x)\) by \(g(x)\):

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

So, the result is:

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

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

The domain of the function \(\left( \frac{f}{g} \right)(x)\) is determined by the values of \(x\) for which the function is defined. Since this is a rational function, it is not defined where the denominator is zero:

\[
7 + x \neq 0
\]
\[
x \neq -7
\]

Thus, the domain of \(\left( \frac{f}{g} \right)(x)\) is all real numbers except \(x = -7\).

In interval notation, the domain is:

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

### Summary

- (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 of \(\left( \frac{f}{g} \right)\) is \( (-\infty, -7) \cup (-7, \infty) \)