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 critical points of the function $$f(x) = 6x^2 - x^3$$ and indicate whether they correspond to a local minimum or maximum using the second derivative test. Answer: The critical points are $$x = 0$$ (local minimum) and $$x = 4$$ (local maximum).

Step-by-step solution

Find the first derivative of the function

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

Set the first derivative equal to zero and solve for x

To find possible critical points, set the first derivative equal to zero and solve for $$x$$: $$12x - 3x^2 = 0$$ Factor out $$3x$$: $$3x(4 - x) = 0$$ Now, set each factor equal to zero: $$3x = 0 \Rightarrow x = 0$$ $$4 - x = 0 \Rightarrow x = 4$$ The potential critical points are $$x = 0$$ and $$x = 4$$.

Find the second derivative of the function

Now find the second derivative by taking the derivative of the first derivative: $$f''(x) = \frac{d(12x - 3x^2)}{dx} = 12 - 6x$$

Use the Second Derivative Test

Evaluate the second derivative at the critical points $$x = 0$$ and $$x = 4$$: $$f''(0) = 12 - 6(0) = 12$$ Since $$f''(0) > 0$$, the function has a local minimum at $$x = 0$$. $$f''(4) = 12 - 6(4) = -12$$ Since $$f''(4) < 0$$, the function has a local maximum at $$x = 4$$. The critical points are $$x = 0$$ (local minimum) and $$x = 4$$ (local maximum).

Key concepts

First Derivative

Understanding the first derivative of a function is critical for analyzing the behavior of the function's graph. It serves as the cornerstone for finding critical points, which are points where the graph of the function could change direction.

To find the first derivative, we use rules of differentiation such as the power rule, which states that the derivative of a function like \( x^n \) is \( nx^{n-1} \). In our exercise, applying the power rule to \( f(x) = 6x^2 - x^3 \) gives us the first derivative \( f'(x) = 12x - 3x^2 \).

Setting the first derivative equal to zero is the method we use to find potential critical points, where the function could have a local maximum or minimum. This is because the zeros of the first derivative indicate where the slope of the tangent to the function's graph is horizontal. In our example, this leads us to find the critical points at \( x = 0 \) and \( x = 4 \).

Second Derivative Test

Once critical points are found using the first derivative, the second derivative test helps us determine the concavity of the function at those points. This test involves taking the second derivative of our function and plugging in the critical points.

The rule of thumb is simple: if the second derivative at a critical point is positive, the function is concave up at that point, suggesting a local minimum. If it's negative, the function is concave down, indicating a local maximum. A zero second derivative, however, may require further testing, as it is inconclusive on its own.

In our exercise, we calculated that \( f''(x) = 12 - 6x \). Evaluating this at our critical points, we determined that at \( x = 0 \), the second derivative is positive, indicating a local minimum, and at \( x = 4 \), it is negative, indicating a local maximum. This test gives us a quick and powerful way to classify critical points without having to fully analyze the graph of the function.

Local Maximum and Minimum

Local maxima and minima refer to points on a graph where the function reaches a peak or a valley within a specific range. They are often found at critical points, where the first derivative of the function equals zero or is undefined.

A local maximum is the highest point in the immediate area, whereas a local minimum is the lowest point. From a practical viewpoint, finding these local extrema can be crucial - for instance, when determining the most efficient use of resources or when calculating maximum profit or minimum cost in economics.

In our exercise example, the critical points gave us a local minimum at \( x = 0 \) and a local maximum at \( x = 4 \). This information tells us about the nature of the graph of \( f(x) \), and if we were to graph it, we would see a valley at \( x = 0 \) and a peak at \( x = 4 \), helping us visualize the behavior of the function in those regions.