Problem 23 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(g(x)=x^{2}(6-x)^{3}\)

Step-by-Step Solution

Answer

The given function \(g(x) = x^{2}(6-x)^{3}\) has a relative minimum at \(x=0\). The result at \(x=6\) from the Second Derivative Test was inconclusive.

Step 1: Find the First Derivative of the Function

We first need to find the derivative of \(g(x)\). To do this, we will use the Product Rule which states that \((uv)' = u'v + uv'\). Here, \(u=x^{2}\) and \(v=(6-x)^{3}\), so \(g'(x) = 2x(6-x)^{3} + x^{2}\cdot 3(6-x)^{2}\cdot (-1) = 2x(6-x)^{3} - 3x^{2}(6-x)^{2}\).

Step 2: Find the Critical Points

Critical points occur where the derivative is zero or undefined. In this case, \(g'(x)\) is only undefined at \(x = 6\), but we can also set \(g'(x) = 0\) to find any other potential critical points. Doing so gives us \(2x(6-x)^{3} - 3x^{2}(6-x)^{2} = 0\). Simplifying this yields \(x = 0, 6\).

Step 3: Find the Second Derivative of the Function

Now that we have our critical points, we need to find the second derivative of \(g(x)\) to be able to use the Second Derivative Test. Just like we used the Product Rule to calculate the first derivative, we'll do so again to calculate the second derivative, \(g''(x)\). The second derivative in this case is a little bit complicated, but it simplifies to \(g''(x) = 12(6-x)^{2} - 28x(6-x)\).

Step 4: Use the Second Derivative Test

Next, we use the Second Derivative Test, which says that if a function's second derivative at a point is positive, the function has a relative minimum there, if it's negative, the function has a relative maximum there. Evaluating \(g''(x)\) at \(x=0\), we get \(12(6)^{2} = 432 > 0\), so \(x=0\) is a relative minimum. Evaluating \(g''(x)\) at \(x=6\), we get \(0\), which is neither positive nor negative, so the test is inconclusive at this point.