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 if there is a local maximum or minimum at the critical point of the function $$f(x) = 4-x^{2}$$, and find the corresponding x-value. Answer: There is a local maximum at x = 0 for the function $$f(x)=4-x^{2}$$.

Step-by-step solution

Find the first derivative

First, find the derivative of the given function $$f(x) = 4-x^{2}$$ with respect to x: $$f'(x) = -2x$$

Locate critical points

A critical point occurs where the derivative is equal to zero or undefined. The derivative $$f'(x) = -2x$$ is never undefined, so we can set it equal to zero and solve for x: $$-2x = 0 \implies x = 0$$ So, there is one critical point at x = 0.

Find the second derivative

To apply the Second Derivative Test, we need to find the second derivative of $$f(x)=4-x^{2}$$: $$f''(x) = -2$$

Apply the Second Derivative Test

The second derivative is a constant value, $$f''(x)=-2$$, which means it has the same sign at every point. Since it is negative, the function is concave down, and the critical point at x = 0 corresponds to a local maximum. So, using the Second Derivative Test, we can conclude that there is a local maximum at x = 0 for the function $$f(x)=4-x^{2}$$.

Key concepts

Second Derivative Test

The Second Derivative Test is a handy tool used to determine the nature of critical points identified by the first derivative. Once you've pinpointed these critical points by setting the first derivative to zero, the second derivative comes into play. It involves the following:

• If the second derivative is positive at a critical point, the function is concave up at that point. This indicates a local minimum.
• If the second derivative is negative at a critical point, the function is concave down at that point, indicating a local maximum.
• If the second derivative equals zero, the test fails, meaning it doesn’t provide any information about that critical point.

For the function given, \(f(x) = 4 - x^2\), the second derivative is \(f''(x) = -2\), a constant negative value. This tells us the function is concave down at all points, confirming any critical point as a local maximum.

First Derivative

The first derivative of a function is the rate at which the function's value changes with respect to changes in the input value \(x\). It is represented as \(f'(x)\) and is crucial for locating critical points.

For our function \(f(x) = 4 - x^2\), the first derivative is \(f'(x) = -2x\). By solving \(f'(x) = 0\), we find the critical points—where the slope of the tangent to the curve is zero, indicating potential local maxima, minima, or points of inflection. In this case, \( -2x = 0 \) gives \( x = 0 \), our single critical point.

Concavity

Concavity relates to the way a function curves. A function is concave up if it looks like a cup (\(^\cup\)) and concave down if it looks like a cap (\(^\cap\)). This concept is closely connected to the second derivative.

• If the second derivative, \(f''(x)\), is positive, the function is concave up at that point.
• If \(f''(x)\) is negative, the function is concave down.

For \(f(x) = 4 - x^2\), we calculated \(f''(x) = -2\), indicating the function is concave down everywhere. This influences the shape of the graph, making it a downward parabola.

Local Maximum

A local maximum is a point where the function value is higher than all other nearby points. It represents a peak of the curve within a particular interval.

For \(f(x) = 4 - x^2\), we determined \(x = 0\) as a critical point where the first derivative is zero. The second derivative test showed \(f''(x) = -2\), confirming the function is concave down and thus pointing to a local maximum. At \(x = 0\), \(f(0) = 4\), thus the local maximum value is 4.