91Ó°ÊÓ

To determine the relative extrema of a function, we first need to identify its critical points. These are specific points on the graph where the function's derivative is zero or undefined. In simpler terms, critical points help us find where the function might have turning points. For the function \( f(x) = x^4 - 2x^3 \), we start by finding the first derivative, \( f'(x) = 4x^3 - 6x^2 \). This derivative represents the slope of the tangent to the curve at any point \( x \).

We set this first derivative equal to zero to find the critical points:

• \( 4x^3 - 6x^2 = 0 \)
• Factor to get: \( 2x^2(2x - 3) = 0 \)
• This gives critical points at \( x = 0 \) and \( x = \frac{3}{2} \).

Both points are potential candidates for relative extrema. The next steps will help us determine whether these are indeed maximas or minimas.

Once we have the critical points, the First Derivative Test comes in handy to classify these points as relative maxima or minima. This test examines the sign of the derivative on either side of each critical point, which gives insight into the slope behavior.

For the point \( x = 0 \):

• Check intervals \(-\infty < x < 0\) and \(0 < x < \frac{3}{2}\).
• For \(-\infty < x < 0\): \( f'(x) > 0 \), slope is positive (function is increasing).
• For \(0 < x < \frac{3}{2}\): \( f'(x) < 0 \), slope is negative (function is decreasing).

Similarly, check around \( x = \frac{3}{2} \):

• For \(x > \frac{3}{2}\), \( f'(x) > 0 \), indicating the function starts to increase again.

These observations show that \( x = 0 \) is neither a maxima nor a minima, while \( x = \frac{3}{2} \) is a relative minimum.

For further confirmation of the nature of our critical points, we can employ the Second Derivative Test. This test offers a more straightforward approach by looking at the concavity of the function around critical points.

The second derivative for our function is \( f''(x) = 12x^2 - 12x \). We plug in our critical points:

• At \( x = 0 \), \( f''(0) = 12(0)^2 - 12(0) = 0 \). The test is inconclusive here since \( f''(0) = 0 \) tells us nothing about concavity.
• At \( x = \frac{3}{2} \), \( f''\left(\frac{3}{2}\right) = 12\left(\frac{3}{2}\right)^2 - 12\left(\frac{3}{2}\right) = \frac{27}{2} - 18 \).

Since \( f''\left(\frac{3}{2}\right) > 0 \), the graph is concave up at this point, indicating a relative minimum. This aligns with what we found using the First Derivative Test. Therefore, \( x = \frac{3}{2} \) is a certain relative minimum, reaffirming our earlier results.