Question

sales of a new model of digital camera are approximated by s(x)=4700 - 3800e^{-x}, where x represents the number of years that the digital camera has been on the market, and s(x) represents sales in thousands of dollars. (a) find the sales in year 0. (b) when will sales reach $4,511,000? (c) find the limit on sales.

Explanation:

Step1: Substitute x = 0 for part (a)

Substitute \(x = 0\) into \(S(x)=4700 - 3800e^{-x}\). Since \(e^{0}=1\), we have \(S(0)=4700-3800\times1\).
\[S(0)=4700 - 3800=900\]

Step2: Solve for x in part (b)

Set \(S(x) = 4511\) (since \(S(x)\) is in thousands of dollars). So, \(4511=4700 - 3800e^{-x}\). First, rearrange the equation:
\[3800e^{-x}=4700 - 4511=189\]
Then, \(e^{-x}=\frac{189}{3800}\). Take the natural - logarithm of both sides: \(-x=\ln(\frac{189}{3800})\), and \(x =-\ln(\frac{189}{3800})=\ln(\frac{3800}{189})\approx\ln(20.1058)\approx3\)

Step3: Find the limit in part (c)

We know that \(\lim_{x
ightarrow+\infty}e^{-x}=0\). So, \(\lim_{x
ightarrow+\infty}S(x)=\lim_{x
ightarrow+\infty}(4700 - 3800e^{-x})\).
\[ \lim_{x
ightarrow+\infty}S(x)=4700-3800\times0 = 4700\]

Answer:

(a) 900
(b) Approximately 3 years
(c) 4700