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. $$f(x)=2 x^{3}-3 x^{2}+12$$

Short answer

Answer: The function has a local maximum at $$x = 0$$ and a local minimum at $$x = 1$$.

Step-by-step solution

First Derivative

To find the first derivative of the function $$f(x) = 2x^3 - 3x^2 + 12$$, apply the power rule for each term, which results in: $$f'(x) = 6x^2 - 6x$$ Step 2: Find the critical points

Critical Points

To find the critical points, we will set the first derivative equal to zero and solve for x: $$6x^2 - 6x = 0$$ Factor out $$6x$$: $$6x(x-1) = 0$$ Thus, the critical points are $$x = 0$$ and $$x = 1$$. Step 3: Find the second derivative

Second Derivative

To find the second derivative of the function, differentiate the first derivative: $$f''(x) = 12x - 6$$ Step 4: Use the Second Derivative Test

Second Derivative Test

Now, we'll use the second derivative test to classify the critical points as local maxima, local minima, or neither. Plug the critical points into the second derivative: For $$x = 0$$: $$f''(0) = 12(0) - 6 = -6 < 0$$ Since $$f''(0)$$ is negative, the graph is concave down and $$x = 0$$ is a local maximum. For $$x = 1$$: $$f''(1) = 12(1) - 6 = 6 > 0$$ Since $$f''(1)$$ is positive, the graph is concave up and $$x = 1$$ is a local minimum. In conclusion, the function $$f(x) = 2x^3 - 3x^2 + 12$$ has a local maximum at $$x = 0$$ and a local minimum at $$x = 1$$.

Key concepts

Critical Points

Critical points are crucial in determining where a function's graph changes its trend. These are the x-values where the function's first derivative is zero or undefined. In our exercise, the first step to finding critical points for the function f(x) = 2x^3 - 3x^2 + 12 is to calculate its first derivative, f'(x). Once calculated as f'(x) = 6x^2 - 6x, setting it equal to zero and solving for x reveals that the critical points are at x = 0 and x = 1. These points correspond to where the slope of the tangent is zero, indicating potential locations for local maxima or minima or points of inflection.

Identifying critical points can signal where to investigate the function further for maximum and minimum values, which is essential in optimization problems and understanding the function's behavior.

Concavity

Concavity refers to how a function curves. If the function curves upwards like a cup, it is said to be 'concave up,' representing an area where the function has positive curvature. On the other hand, if the function curves downwards like a frown, it is 'concave down,' indicating negative curvature. To determine concavity, one must inspect the second derivative of the function, f''(x). In the given function, the second derivative is f''(x) = 12x - 6.

When we plug the critical points into the second derivative, as shown in the exercise, we find that the function is concave down at x=0 and concave up at x=1. Understanding concavity is valuable because it helps predict the behavior of the function and the nature of the critical points, which is instrumental when sketching graphs and analyzing the function.

First Derivative

The first derivative of a function, often denoted as f'(x), represents the rate of change or the slope of the tangent line at any point on the graph of the function. Obtaining the first derivative is a fundamental concept in calculus, typically using rules like the power rule, product rule, or chain rule. In our example, the first derivative of f(x) = 2x^3 - 3x^2 + 12 was found using the power rule. The resulting derivative, f'(x) = 6x^2 - 6x, is used to identify the critical points by setting f'(x) to zero.

In a graph, where f'(x) > 0, the function is increasing, and where f'(x) < 0, the function is decreasing. These changes give insight into the function's nature and allow the prediction of key features such as local maxima and minima.

Local Maxima and Minima

Local maxima and minima represent the highest and lowest points, respectively, in a small region around a point on the graph. They are essential in analyzing the extremities of a function. Using the Second Derivative Test, we can classify critical points based on the concavity at those points.

For the function in the exercise, the second derivative test indicates that at the critical point x=0, where f''(x) < 0, there is a local maximum since the graph is concave down at that point. Conversely, at x=1, where f''(x) > 0, there is a local minimum with the graph's concavity facing upwards. This test is a powerful tool for pinpointing extrema and is particularly useful when dealing with applications involving optimization and curve sketching.