Sketch the curve of the function

     f(t) = -t³ + 5t² - 3t + 5

and show all the significant features on it.

It is given that the life cycle of Auxin is 7 weeks, so the domain of the function is (0, 7).

Setting x equals 0, y intercept can be found. That is

     f(0) = 5

Setting f(x) = 0, x intercept can be calculated. That is

     -t³ + 5t² - 3t + 5 = 0

Since this equation involves a high order polynomial function, it is hard to calculate. However, an estimated value of 5 weeks can be determined and used for preliminary plotting.

Substituting -x into the function to determine the odevity of the cell elongation speed function and gives

     f(-x) = -(-t)³ + 5(-t)² - 3(-t) + 5
            = t³ + 5t² + 3t + 5

Since f(-x) ≠ f(x) and f(-x) ≠ -f(x), this function is not symmetric with respect to origin or y axis.

Next, calculate the first derivative to determine the monotonocity of the function. That is

     df(x)/dx = -3t² + 10t - 3
                 = -(3t - 1)(t - 3)

The critical points occurs when the derivative equals 0, which is at 1/3 and 3. The domain of time is between 0 and 7, which is broken down to three intervals by 1/3 and 3, they are (0, 1/3), (1/3, 3), and (3, 7). The sign and the monotonicity are shown in the table. A "+" indicates a positive expression and a "-" means a negative expression.

The local extreme values can be calculated at the critical points. They are

     f(1/3) = -(1/3)³ + 5(1/3)² - 3(1/3) + 5 = 122/7 = 4.5

     f(3) = -(3)³ + 5(3)² - 3(3) + 5 = 14

Therefore, the local minimum value is 4.5 and the local maximum value is 14.

In order to analyze the concavity and find the inflection point, the second derivative is needed. That is

     d²f(x)/dx² = d(-3t² + 10t - 3)/dx = -6t + 10

Thus, the inflection point is at ( 1.7, 9.3)

Therefore,

The curve is concave upward on (0, 5/3) and concave downward on (5/3, 7).

According to all the information listed above, the curve of the function f(t) = -t³ + 5t² - 3t + 5 is sketched on the left with all the significant features.