Problem 39 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=\sqrt{16-x^{2}} \end{equation}

Step-by-Step Solution

Answer

Local and absolute maximum at (0, 4); absolute minima at (-4, 0) and (4, 0). No inflection points.

Step 1: Determine the Domain

The function is defined for \( y = \sqrt{16-x^2} \) as long as the expression under the square root is non-negative. Thus, \( 16 - x^2 \geq 0 \). Solving for \( x \), we get \( -4 \leq x \leq 4 \). Therefore, the domain is \( -4 \leq x \leq 4 \).

Step 2: Find Critical Points

To identify critical points, compute the derivative. Start with the function \( y = \sqrt{16-x^2} \), or equivalently \( y = (16-x^2)^{1/2} \). Use the chain rule to find \( y' = \frac{d}{dx}(16-x^2)^{1/2} = \frac{-x}{\sqrt{16-x^2}} \). Set \( y' = 0 \) to find critical points: \( \frac{-x}{\sqrt{16-x^2}} = 0 \), which gives \( x = 0 \).

Step 3: Determine Local Extrema

Consider the critical point \( x = 0 \). Evaluate the second derivative: \( y'' = \frac{-16}{(16-x^2)^{3/2}} \). At \( x = 0 \), \( y'' = -\frac{1}{8} < 0 \), thus \( x = 0 \) is a local maximum. Since the derivative changes from positive to negative at this point, \( x = 0 \) is also an absolute maximum due to the symmetry of the function throughout its domain.

Step 4: Identify Endpoint Behavior

Evaluate the function at the endpoints \( x = -4 \) and \( x = 4 \). At these points, \( y(4) = y(-4) = \sqrt{0} = 0 \). These represent absolute minima, as the function is non-negative within its domain.

Step 5: Determine Inflection Points

Inflection points occur where the second derivative changes sign. However, \( y'' = \frac{-16}{(16-x^2)^{3/2}} \) remains negative over the entire domain \( -4 < x < 4 \), meaning that there are no inflection points within the domain.

Step 6: Graph the Function

The graph of \( y = \sqrt{16-x^2} \) is a semicircle with radius 4 centered at the origin, spanning from \( x = -4 \) to \( x = 4 \), with the highest point at \( (0, 4) \). The absolute minimum points are at \( (-4, 0) \) and \( (4, 0) \). Its geometric shape confirms the calculations of local and absolute extrema.

Key Concepts

Critical Points

In calculus, a critical point occurs where the first derivative of a function is zero or undefined. These points are important because they can indicate possible locations of local extrema — points where the function reaches local maximum or minimum values.
To find critical points in the function \( y = \sqrt{16-x^2} \), we first compute the derivative, \( y' = \frac{-x}{\sqrt{16-x^2}} \). By setting \( y' = 0 \), we solve for \( x \) to find \( x = 0 \) as a critical point. This calculation means at \( x = 0 \), the slope of the tangent to the curve is horizontal, and it is a candidate for being a peak or valley. In this situation, because it is a semicircle, the critical point is indeed a peak.

Local Extrema

Local extrema refers to the highest or lowest values found in a small "neighborhood" around a point on a curve. These are identified at critical points and can be found using the second derivative test.
With our function \( y = \sqrt{16-x^2} \), evaluating the second derivative \( y'' = \frac{-16}{(16-x^2)^{3/2}} \), we find \( y'' < 0 \) at the critical point \( x = 0 \), which implies a local maximum. A negative second derivative indicates the function is concave down at this point, confirming that \( x = 0 \) is a local maximum for the semicircle.

Absolute Extrema

Absolute extrema are the largest or smallest values a function takes on over its entire domain. These include both endpoints and any critical points within the domain.
For \( y = \sqrt{16-x^2} \), the symmetry and geometry of the semicircle suggest the maximum point is at \( x = 0 \) with \( y = 4 \), which is the absolute maximum. The endpoints \( x = -4 \) and \( x = 4 \) both yield \( y = 0 \), the absolute minimum values of the semicircle. Thus, the domain endpoints play a crucial role in determining absolute extrema.

Domain

The domain of a function defines the set of possible input values (\( x \) values) for which the function is defined. In the function \( y = \sqrt{16-x^2} \), the expression under the square root must be non-negative, leading to \( 16 - x^2 \geq 0 \).
Solving this inequality gives \( -4 \leq x \leq 4 \). Hence, the function is defined and will produce real numbers only within this interval along the \( x \) axis. This restriction means the graph forms a semicircle, as the function is only real for \( x \) values within the domain specified.

Inflection Points

Inflection points are points on the curve where the curvature changes direction, or the concavity switches between concave up and concave down. To find these, we examine where the second derivative changes sign.
For \( y = \sqrt{16-x^2} \), the second derivative is \( y'' = \frac{-16}{(16-x^2)^{3/2}} \). Within the domain \( -4 < x < 4 \), this value remains negative, implying the function is consistently concave down throughout. Consequently, there are no inflection points in this domain, as there's no change from concave down to concave up or vice versa.