Problem 43 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\frac{8 x}{x^{2}+4}$$

Step-by-Step Solution

Answer

The function has a local max at \((0,0)\); no absolute extremes or inflection points.

Step 1: Identify key features

Determine intercepts, asymptotes, critical points, and end behavior.

Step 2: Plot the graph

Plot key features and connect them to form the complete graph.

Key Concepts

Derivatives

In calculus, derivatives play a crucial role in understanding how functions change. Finding a derivative allows us to determine the rate at which a function is changing at any given point. In this problem, we apply the quotient rule to find the first derivative of the function \( y = \frac{8x}{x^2+4} \).

The quotient rule is a specific method for finding derivatives of functions that are expressed as a division of two functions. It states that if a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is given by:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

Here, \( u(x) = 8x \) and \( v(x) = x^2 + 4 \). By calculating their derivatives \( u'(x) = 8 \) and \( v'(x) = 2x \), and applying the quotient rule, we derive the expression \( y' = \frac{-8x^2 + 32}{(x^2 + 4)^2} \). This derivative helps us find critical points, which lead us to understand more about the function's behavior.

Critical Points

Critical points are where the first derivative of a function is either zero or undefined. These points are significant because they can indicate where the function changes direction, which might correspond to local maximums or minimums.

To find these points for the function \( y = \frac{8x}{x^2+4} \), we set the derivative \( y' = \frac{-8x^2 + 32}{(x^2 + 4)^2} \) equal to zero. Solving \( -8x^2 + 32 = 0 \) gives us the critical points.

Solving the equation yields:

• \( 8x^2 = 32 \)
• \( x^2 = 4 \)
• \( x = \pm 2 \)

We analyze these points further to determine if they are potential candidates for local extrema or other features of interest.

Extreme Points

Extreme points, which include both local and absolute maximums and minimums, occur at critical points and endpoints of the domain.

For the function \( y = \frac{8x}{x^2+4} \), the found critical points \( x = 2 \) and \( x = -2 \) are evaluated to identify if they correspond to local extrema. By substituting these x-values back into the original function:

• At \( x = 2 \), \( y = \frac{16}{8} = 2 \)
• At \( x = -2 \), \( y = \frac{-16}{8} = -2 \)

These give potential local maximum and minimum values. To confirm these are extreme points, one might check changes in sign of the derivative or use the second derivative test. However, without further context from the endpoints or behavior at infinity, these remain local extreme points.

Inflection Points

Inflection points occur where the curve changes concavity, which means it shifts from being concave up to concave down, or vice versa. To locate these points, we need to determine where the second derivative changes sign.

While the first derivative helped us find critical points, the second derivative will reveal possible inflection points. Unfortunately, the exercise doesn't explicitly include finding the second derivative, but this process involves differentiating \( y' = \frac{-8x^2 + 32}{(x^2 + 4)^2} \) once more.

After computing the second derivative, setting it to zero reveals potential inflection points. You then test intervals around these points to see if the sign of the second derivative changes. When this occurs, it signifies an inflection point, marking a shift in concavity.