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. $$f(x)=4-x^{2}$$

Short answer

Question: Determine the local maximum of the function $$f(x) = 4 - x^2$$. Answer: There is a local maximum at x = 0.

Step-by-step solution

Find the first derivative

To do this, we will apply the power rule to the given function: $$f(x) = 4 - x^2$$ $$f'(x) = -2x$$

Set the first derivative equal to zero and solve for x

We set the first derivative equal to zero and solve for x to find the critical points. $$-2x=0$$ $$x=0$$ There is one critical point, x = 0.

Determine the second derivative

Now, we will find the second derivative of the function by applying the power rule to the first derivative. $$f'(x)=-2x$$ $$f''(x)=-2$$

Evaluate the second derivative at each critical point

We evaluate the second derivative at the critical point x = 0. $$f''(0)=-2$$ The second derivative is negative, which indicates that the critical point x = 0 corresponds to a local maximum. Thus, the given function has a local maximum at x = 0.

Key concepts

Second Derivative Test

Understanding the second derivative test is essential for identifying the nature of critical points in a function, which in turn tells us about the function's concavity and potential local maxima and minima. In calculus, after finding the critical points, we don't just want to know where the function's slope is zero, we also want to understand the behavior of the function near these points.

Here's a simple way to use the second derivative test: after determining the function's critical points by setting the first derivative equal to zero, we calculate the second derivative. When we plug the critical points into the second derivative, a negative value suggests that the graph of the function is concave down at that point, indicating a local maximum. Meanwhile, a positive value means the graph is concave up, signaling a local minimum. If the second derivative is zero, the test is inconclusive, and we may need to use other methods to characterize the critical point.

In our exercise, the second derivative at the critical point x=0 is -2. Since this is negative, it confirms that at x=0, the function has a local maximum according to the second derivative test.

Local Maximum and Minimum

Local maximum and minimum points on a graph are the 'peaks' and 'valleys' where a function reaches its highest or lowest value, respectively, within a localized interval. These points are of interest because they can tell us a lot about the function's overall behavior and are often key when solving optimization problems in calculus.

A local maximum is found at a point in the domain where the function's value is higher than all other values in its immediate vicinity. Conversely, a local minimum is where the function's value is lower than any other nearby points. It's important to distinguish local extrema from global ones, as local maxima and minima are not necessarily the highest or lowest points on the entire graph.

In the provided exercise, after we've applied the second derivative test, we identified a local maximum at the critical point. Since derivatives represent rates of change, the transition from increasing to decreasing at a critical point (as indicated by the sign of the second derivative) helps isolate these local extrema. This understanding allows us to better analyze and comprehend the function's progression.

Power Rule Differentiation

The power rule is a fundamental concept in differential calculus, a quick and efficient way to find the derivative of a function that involves a power of x. In general, for any function of the form \( f(x) = ax^n \), where \( a \) and \( n \) are constants, the power rule states that the derivative of \( f \) with respect to \( x \) is \( f'(x) = anx^{n-1} \). It streamlines the differentiation process, allowing us to easily find critical points and hence analyze the function's behavior.

In our exercise, we applied the power rule to differentiate the function \( f(x) = 4 - x^2 \). This rule tells us to bring down the exponent as a coefficient and then decrease the exponent by one. Therefore, the derivative is \( f'(x) = -2x^{2-1} \) simplifying to \( f'(x) = -2x \). Such a straightforward method allows us to quickly get to the essence of the problem: analyzing the function's critical points and determining their nature in the greater context of the function's graph.