Distance increasing question, please HELP?

Question

The position of a car from a fixed point is modeled by f(t) = 2t^3 – 21t^2 + 60t, t ≥ 0, where t is measured in minutes and position is measured in meters. When is the car’s distance increasing from the fixed point?

Answer 1

The distance is increasing when the derivative is greater than 0.

f'(t) = 6t^2 - 42t + 60 = 6(t^2 - 7t + 10) = 6(t-5)(t-2)
This will be greater than 0 whenever either both terms are positive or both terms are negative.
Thus, it is positive whenever t>5 or t<2.

Answer 2

Distance is increasing when slope of distance function is positive, that is, when df/dt > 0, so

f'(t) = 6t² - 42t + 60 = 0
t² - 7t + 10 = 0
(t-2)(t-5) = 0
t = 2, t = 5 are the critical points.
f(t) increases on (-∞,2) and (5,∞), decreases on (2,5)

Answer 3

f(t) = 0
2t^3 - 21t^2 + 60t = 0
2t (t^2 - 21/2 t + 30) = 0
2t (t^2 - 21/2 t + (21/4)^2 - (21/4)^2 + 30) = 0
2t((t- 21/4)^2 - (441 - 480)/16) = 0
2t ((t - 21/4)^2 + 39/16) = 0

t = 0

Answer 4

Set f(t) = 0. Solve for the cubic. You'll get three answers, one of which will be greater than 0. That's your answer.

Answer 5

You have to factorize your equation and draw up a head sign schema (It probably has a better name. Can't remember the english name atm, soz)

You factorize by applying the standard solver for second order polynomials