### Approach with Derivatives

Recall equation (1)

 W = 0.03t2 - 0.39t + 7.3 (1)

Obtain the first derivative dW/dt by differentiating the equation on both sides, i.e.

 dW/dt = 0.03*2t - 0.39 (2)

In relation to the graph W vs t given by (1), dW/dt at a t-value will give the slope of the tangent at that t-value. What will be the slope of the tangent at the point where a graph has a minimum? Of course, it is zero (if you are not sure about this, review the section on slope). Therefore, in (2), if we make

 dW/dt = 0

one obtains

 0.03*2t - 0.39 = 0 (3)

Now, by solving (3), we can find the t-value

 t = 0.39/(2*0.03) = 6.5 (4)

At this juncture, let us pause for a minute. Do we know for sure that this t-value corresponds to the minimum value of W? No, we do not! This t-value may very well correspond to the maximum value for W, because the slope of the tangent at the point where a graph has a maximum is also zero!!

For the case study presented in this lesson, show that the t-value in (4) corresponds to a minimum.

So, the minimum weight will occur 6.5 days after the baby is born and that minimum weight will be

 W = 0.03*(6.5)2 - 0.39*(6.5) + 7.3

This means that the baby will start to grow after 6 and a 1/2 days from birth.

Question: Could we have found this result without using the concept "derivative"? The answer is yes. See how with the derivative-free and algebraic approaches.