Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=6 x^{2}-x^{3}$$

Short answer

Question: Determine the local maxima and local minima of the function $$f(x) = 6x^2 - x^3$$. Answer: The function has a local minimum at x=0 and a local maximum at x=4.

Step-by-step solution

Find the first derivative

To find the first derivative of the function, apply the power rule: $$f'(x) = \frac{d}{dx}(6x^2 - x^3) = 12x - 3x^2$$

Find the critical points

A critical point occurs when the first derivative is equal to zero or undefined. In this case, f'(x) is a polynomial, so it's never undefined. Set f'(x) to zero and solve for x: $$12x - 3x^2 = 0$$ $$x(12 - 3x) = 0$$ This equation has two solutions: x=0 and x=4. These are the critical points.

Find the second derivative

To find the second derivative of the function, differentiate f'(x) with respect to x: $$f''(x) = \frac{d^2}{dx^2}(12x-3x^2)= 12 - 6x$$

Apply the Second Derivative Test

Using the critical points found in Step 2, evaluate the second derivative at these points to determine concavity and classify them as local maxima, local minima, or neither. For x=0: $$f''(0) = 12- 6(0) = 12$$ Since f''(0) > 0, the function is concave up at x=0, which means there is a local minimum at x=0. For x=4: $$f''(4) = 12 - 6(4) = -12$$ Since f''(4) < 0, the function is concave down at x=4, which means there is a local maximum at x=4. In conclusion, there is a local minimum at x=0 and a local maximum at x=4.

Key concepts

First Derivative

The first derivative of a function is foundational in calculus, especially for identifying critical points. When we say "first derivative," we are looking at how a function changes, or slopes, at any given point. To find it, we often use calculus rules, like the power rule: if you have a term like \(ax^n\), the derivative becomes \(n \cdot ax^{n-1}\).
For example, with the function \(f(x)=6x^2-x^3\), it becomes \(f'(x) = 12x - 3x^2\).

So why is the first derivative important? It helps us find **critical points**, where the derivative equals zero or becomes undefined. These points are where the function's slope changes direction, potentially indicating a peak or valley in the graph.
For \(f(x)\), the equation \(12x - 3x^2 = 0\) can be solved by factoring:\
\[x(12 - 3x) = 0\] leading us to critical points \(x = 0\) and \(x = 4\).

These points mark where the function's behavior changes, which is why they are crucial in analyzing local maxima and minima.

Second Derivative Test

After pinpointing critical points using the first derivative, the second derivative becomes an essential tool. The second derivative, denoted as \(f''(x)\), examines the curvature or concavity of a function.
For our function \(f(x)=6x^2-x^3\), the second derivative oversees how the graph bends:\
Suppose \(f'(x) = 12x - 3x^2\) became \(f''(x) = 12 - 6x\).

Now, the **Second Derivative Test** reveals if each critical point is a local maximum or a local minimum:

• If \(f''(x) > 0\), the curve opens upwards, indicating a local minimum.
• If \(f''(x) < 0\), the curve opens downwards, suggesting a local maximum.
• If \(f''(x) = 0\), the test is inconclusive; other methods might be needed.

In our example, at \(x = 0\), \(f''(0) = 12\) means the graph is concave up, pointing to a local minimum.
Whereas at \(x = 4\), \(f''(4) = -12\) shows the graph as concave down, highlighting a local maximum.

This test efficiently classifies the nature of each critical point, making it a powerful step in function analysis.

Local Maxima and Minima

Understanding local maxima and minima helps in grasping the peaks and valleys within a function. A local maximum is where a function takes a larger value than nearby points, appearing as a peak.
In contrast, a local minimum is a point where the function's value is smaller than those close by, forming a valley.

By utilizing the first and second derivatives, we can determine these points:

• First, critical points are found where the slope (first derivative) is zero.
• Then, the second derivative test determines the curvature at these points to classify them.

In our specific example, for the function \(f(x)=6x^2-x^3\):
We identified critical points at \(x = 0\) and \(x = 4\).
Applying the second derivative test:

• At \(x = 0\), \(f''(0) = 12\), yielding a local minimum since the graph curves upwards.
• At \(x = 4\), \(f''(4) = -12\), indicating a local maximum as the graph bends downwards.

This comprehensive approach ensures a clear understanding of the function's behaviors at specific points, which is invaluable for graphing and analysis.