At time t = 0, a seed is planted. After t weeks, the height of the plant is given by f(t) = 0.3t² = 0.6t = 0.5 inches. At what rate is the plant growing after

incomplete posting, and you probably meant + instead of =

but ....

f ' (t) = .6t^2 + .6

whatever is given for t, sub it into the last expression

Actually it would be:

f ' (t) = 0.6t + 0.6
(not t squared as the original reply showed)

To find the rate at which the plant is growing, we need to calculate the derivative of the height function f(t) with respect to time (t). In this case, we have a height function f(t) = 0.3t^2 + 0.6t + 0.5.

To take the derivative of f(t) with respect to t, we can apply the power rule and sum/difference rule of differentiation. The power rule states that the derivative of t^n is n*t^(n-1), where n is a constant.

Applying the power rule to the terms in f(t), we get:

f'(t) = (d/dt)(0.3t^2) + (d/dt)(0.6t) + (d/dt)(0.5)

Using the power rule, the derivatives of the first two terms are:

(d/dt)(0.3t^2) = 0.3 * 2t = 0.6t
(d/dt)(0.6t) = 0.6

The derivative of a constant (0.5 in this case) is zero, so it does not contribute to the rate of growth.

Therefore, f'(t) = 0.6t + 0.6.

To find the rate at which the plant is growing after a specific time t, substitute that value into f'(t). If you provide the specific value of t, I can calculate the rate for you.