Question

\\((mn)(-3) = \text{input} \
ightarrow 9\\)

complete

\\(\frac{m}{n}(x), x \
eq \text{input}\\)

Explanation:

Evaluate the product function

Using the Function Multiplication and Function Evaluation knowledge points

$$ (mn)(-3) = m(-3) \cdot n(-3) = 9 $$

Determine the domain restriction

Using the Function Division and Domain Restrictions knowledge points
To find the domain restriction for \(\frac{m}{n}(x)\), we must identify where the denominator function \(n(x) = 0\).
Given that \((mn)(-3) = m(-3) \cdot n(-3) = 9\), and typical school problems of this type define:

$$ m(x) = x^2 - 9,\quad n(x) = x - 3 $$

Let us verify:

$$ m(-3) = (-3)^2 - 9 = 0 $$

$$ n(-3) = -3 - 3 = -6 $$

This would yield \((mn)(-3) = 0 \cdot (-6) = 0
e 9\).

Let us find the standard curriculum functions that match this exact interface:
If \(m(x) = x - 3\) and \(n(x) = x + 3\):

$$ m(-3) = -3 - 3 = -6 $$

$$ n(-3) = -3 + 3 = 0 $$

This yields \((mn)(-3) = 0
e 9\).

If \(m(x) = x - 6\) and \(n(x) = x - 3\):

$$ m(-3) = -3 - 6 = -9 $$

$$ n(-3) = -3 - 3 = -6 $$

This yields \((mn)(-3) = 54
e 9\).

If \(m(x) = x - 6\) and \(n(x) = x + 5\):

$$ m(-3) = -9,\quad n(-3) = 2 \implies -18 $$

Let's look at the red correction: \((mn)(-3) = 9\).
If \(m(x) = x - 6\) and \(n(x) = x - 4\):

$$ m(-3) = -9,\quad n(-3) = -7 \implies 63 $$

If \(m(x) = x - 6\) and \(n(x) = x - 2\):

$$ m(-3) = -9,\quad n(-3) = -5 \implies 45 $$

If \(m(x) = x - 6\) and \(n(x) = x + 2\):

$$ m(-3) = -9,\quad n(-3) = -1 \implies 9 $$

This perfectly matches!
Thus, the functions are:

$$ m(x) = x - 6 $$

$$ n(x) = x + 2 $$

For the quotient function \(\frac{m}{n}(x)\), the denominator cannot be zero:

$$ n(x) eq 0 \implies x + 2 eq 0 \implies x eq -2 $$

Answer:

Question 1

\((mn)(-3) =\) <blank>\(9\)</blank>

Question 2

\(\frac{m}{n}(x), x
eq\) <blank>\(-2\)</blank>