## Intermediate Algebra: Connecting Concepts through Application

a) $N^{-1}(t) = -\frac{t-4809.8}{315.9}$ b) Domain and range of inverse: Domain: $[ 1019, 5125.7]$ Range: $[-1, 12]$ c) In the year $1999$
$N(t) = -315.9t + 4809.8$ a) Find the inverse Switch the variable $N(t)$ to $y$ and then solve for $y$: Let $N(t) = y$ $y = -315.9t + 4809.8$ Swap the variables $t$ and $y$ to find the inverse: $t = -315.9y + 4809.8$ $t - 4809.8 = -315.9y$ $y = -\frac{t-4809.8}{315.9}$ $N^{-1}(t) = -\frac{t-4809.8}{315.9}$ b) The domain and range of the original function is swapped for the inverse: Domain: $[ 1019, 5125.7]$ Range: $[-1, 12]$ c) Estimate the year $2000 = -315.9t + 4809.8$ $2000 - 4809.8 = -315.9t$ $-2809.8 = -315.9t$ $2809.8 = 315.9t$ $t = 8.89456....$ $t \approx 8.9$ years $t \approx 9 years$ $= 1990 + 9$ $= 1999$