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