Let f(t) = t^3 - 4t^2 + 5t - 2 denote the position of an object (moving in a straight line) at any time t. Find values of t, if any, at which the object is stopped. And apparently this has to be done using limits.
Any help would be appreciated, thanks
Any help would be appreciated, thanks
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You can use the definition a derivative (which uses the limit) and set it equal to zero and solve for t. The derivative of position is velocity, when velocity is zero the object is not moving.
lim (f(t + h) - f(h))/h
h -> 0
Would be the definition of a derivative. Rather than do all of that algebra, I'll just use the power rule (thus, not using limits, so you may want to compute the derivative using the definition)
f '(t) = 3t^2 - 8t + 5
set this equal to zero to find when the object isn't moving.
3t^2 - 8t + 5 = 0
(t - 1)(3t - 5) = 0
t = 1
t = 5/3
lim (f(t + h) - f(h))/h
h -> 0
Would be the definition of a derivative. Rather than do all of that algebra, I'll just use the power rule (thus, not using limits, so you may want to compute the derivative using the definition)
f '(t) = 3t^2 - 8t + 5
set this equal to zero to find when the object isn't moving.
3t^2 - 8t + 5 = 0
(t - 1)(3t - 5) = 0
t = 1
t = 5/3