Answer
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 $1999$
Work Step by Step
$N(t) = -315.9t + 4809.8$
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 $t \approx 9 years$ $= 1990 + 9$ $= 1999$
