Answer:
(a) x = 3 - p²
(b) (0, ∞)
(c) 750 units [ or 0.75 thousand]
Explanation:
Given the equation:
[tex]p^2+x=3[/tex]
Where:
• x = the annual number of units demanded of a product (in thousands).
,
• p = unit price, in dollars per unit.
Part A
[tex]\begin{gathered} p^2+x=3 \\ \text{Subtract }p^2\text{ from both sides.} \\ p^2-p^2+x=3-p^2 \\ \implies x=3-p^2 \end{gathered}[/tex]
x explicitly as a function of p is x=3-p².
Part B
The domain of the function is the set of all possible values of p.
The price is always non-negative, therefore, the practical domain is:
[tex](0,\infty)[/tex]
Part C
At a unit price of 1.50 i.e. p=1.50
[tex]x=3-p^2=3-1.5^2=0.75[/tex]
When the unit price is 1.50, the number of units sold will be:
[tex]\begin{gathered} 0.75\text{ thousands} \\ =0.75\times1000 \\ =750\text{ units} \end{gathered}[/tex]
At a unit price of 1.50, 750 units [ or 0.75 thousand] will be sold.
