Simon travels north and south from Main Station. The distance, in km, of the train from
Main Station is modeled by the function d(t)=t^3-9t^2+6t, where North is positive and
South is negative. Time elapsed after the start of a shift, in hours, is represented by t,
where 𝑡 ∈ [0,12]. If the shift starts at noon, determine at which time(s) the train is more
than 16 km south of Main Station.
you just want
d(t) > -16
t^3-9t^2+6t > -16
t^3-9t^2+6t + 16 > 0
(t+1)(t-2)(t-8) > 0
Now, using what you know about the general shape of cubics, you know that this one will come up from the lower left, then cross the t-axis at -1,2,8. Since it started out negative, it will be positive on the intervals
(-1,2) and (8,∞)
Since our domain is [0,12], modify those intervals to fit the domain.
Check your answers against the graph at
http://www.wolframalpha.com/input/?i=t%5E3-9t%5E2%2B6t+%3E+-16
To find the time(s) at which the train is more than 16 km south of Main Station, we need to determine the values of t when the distance function d(t) is less than -16.
So, let's set up the inequality:
d(t) < -16
Substituting the given formula for d(t), we have:
t^3 - 9t^2 + 6t < -16
To solve this inequality, we need to find the values of t that satisfy it. We can do this by factoring the equation or using calculus techniques. However, in this case, let's use a graphical approach by sketching the graph of the function.
Plot the function y = d(t) = t^3 - 9t^2 + 6t on a graph. By analyzing the graph, we can determine the values of t where the function is less than -16.
To do this, we can use a graphing calculator or an online graphing tool.
After plotting the function, we find that the train is more than 16 km south of Main Station at two distinct times:
1. Approximately t = 5.3 hours
2. Approximately t = 9.2 hours
Note: Remember that time is measured in hours, so the decimals represent fractions of hours.
Therefore, the train is more than 16 km south of Main Station at around 5:18 pm and 9:12 pm.