Questions: Consider the function: f(x) = (81 x^2 + 49) / x Use the First Derivative Test to classify the relative extrema. Write all relative extrema as ordered pairs of the form (x, f(x)). (Note that you will be calculating the values of the relative extrema, as well as finding their locations.)

Solution Steps

To classify the relative extrema using the First Derivative Test, we first need to find the derivative of the function \( f(x) = \frac{81x^2 + 49}{x} \). Then, we set the derivative equal to zero to find the critical points. After identifying the critical points, we analyze the sign of the derivative before and after each critical point to determine whether each point is a relative maximum, minimum, or neither. Finally, we evaluate the function at these critical points to find the corresponding \( f(x) \) values and express the relative extrema as ordered pairs.

Given the function \( f(x) = \frac{81x^2 + 49}{x} \), we first find its derivative. The derivative is:

\[ f'(x) = 162 - \frac{81x^2 + 49}{x^2} \]

To find the critical points, we set the derivative equal to zero:

\[ 162 - \frac{81x^2 + 49}{x^2} = 0 \]

Solving this equation, we find the critical points:

\[ x = -\frac{7}{9}, \quad x = \frac{7}{9} \]

Next, we evaluate the original function \( f(x) \) at each critical point to find the corresponding \( f(x) \) values:

• For \( x = -\frac{7}{9} \):

\[ f\left(-\frac{7}{9}\right) = \frac{81\left(-\frac{7}{9}\right)^2 + 49}{-\frac{7}{9}} = -126 \]

• For \( x = \frac{7}{9} \):

\[ f\left(\frac{7}{9}\right) = \frac{81\left(\frac{7}{9}\right)^2 + 49}{\frac{7}{9}} = 126 \]

Using the First Derivative Test, we determine the nature of each critical point:

• At \( x = -\frac{7}{9} \), the function changes from increasing to decreasing, indicating a relative maximum.
• At \( x = \frac{7}{9} \), the function changes from decreasing to increasing, indicating a relative minimum.

Final Answer