Question

a popular video game retailer develops apps. the total revenue, in dollars per day, after x days can be modeled by the function ( r(x) = 2x^3 + 11x^2 - 23x - 14 ). the total cost, in dollars per day, after x days, can be modeled by the function ( c(x) = 7x^2 + 15x + 26 ). after how many days does the retailer start making money? (\bigcirc) 2 (\bigcirc) 1 (\bigcirc) 5 (\bigcirc) 4

Explanation:

To determine when the retailer starts making money, we need to find when revenue exceeds cost, i.e., \( R(x) > C(x) \). This means \( R(x) - C(x) > 0 \).

Step 1: Calculate \( R(x) - C(x) \)

Subtract the cost function from the revenue function:
\[

$$\begin{align*} R(x) - C(x) &= (2x^3 + 11x^2 - 23x - 14) - (7x^2 + 15x + 26) \\ &= 2x^3 + 11x^2 - 23x - 14 - 7x^2 - 15x - 26 \\ &= 2x^3 + 4x^2 - 38x - 40 \end{align*}$$

\]

Step 2: Test the given values of \( x \) (2, 1, 5, 4) in \( R(x) - C(x) \)

• For \( x = 1 \):

\[

$$\begin{align*} 2(1)^3 + 4(1)^2 - 38(1) - 40 &= 2 + 4 - 38 - 40 \\ &= 6 - 78 \\ &= -72 < 0 \end{align*}$$

\]

• For \( x = 2 \):

\[

$$\begin{align*} 2(2)^3 + 4(2)^2 - 38(2) - 40 &= 16 + 16 - 76 - 40 \\ &= 32 - 116 \\ &= -84 < 0 \end{align*}$$

\]

• For \( x = 4 \):

\[

$$\begin{align*} 2(4)^3 + 4(4)^2 - 38(4) - 40 &= 128 + 64 - 152 - 40 \\ &= 192 - 192 \\ &= 0 \end{align*}$$

\]

• For \( x = 5 \):

\[

$$\begin{align*} 2(5)^3 + 4(5)^2 - 38(5) - 40 &= 250 + 100 - 190 - 40 \\ &= 350 - 230 \\ &= 120 > 0 \end{align*}$$

\]

Answer:

5 (since at \( x = 5 \), \( R(x) - C(x) > 0 \), meaning revenue exceeds cost)