Problem 38 - Calculus of a Single Variable

Question

Using the Second Derivative Test In Exercises \(31-42\) , find all relative extrema. Use the Second Derivative Test where applicable. $$ f(x)=\sqrt{x^{2}+1} $$

Step-by-Step Solution

Answer

The function \(f(x)=\sqrt{x^{2}+1}\) has no relative maximum or minimum.

Step 1: Derive the function

The first derivative of the function \(f(x)=\sqrt{x^{2}+1}\) is given by \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\). Use the power rule and chain rule to calculate the derivative.

Step 2: Find the critical points

Critical points occur where the first derivative is equal to zero or undefined. Set \(f'(x)=0\) and solve for 'x'. However, this equation has no real roots. The derivative is undefined at \(x=\infty\) and \(x= -\infty\), but these aren't in the domain of the function.

Step 3: Find the second derivative

The second derivative of the function is \(f''(x)=\frac{1}{(x^{2}+1)^{3/2}}\). Use the quotient rule and simplification to reach this result.

Step 4: Apply the Second Derivative Test

The Second Derivative Test involves plugging the critical points into the second derivative. In this case, there are no critical points to plug in. However, the second derivative is always positive for any value of 'x', this shows that the graph of the function is concave up everywhere, so there is no relative maximum or minimum.

Key Concepts

Relative Extrema

Relative extrema are the "peaks" and "valleys" of a graph, the points where a function reaches a relative maximum or minimum when compared to nearby points. They occur at the highest or lowest points in a particular section of the graph.

To identify relative extrema, we often use the First and Second Derivative Tests, which help us understand where these peaks and valleys might be.

• A relative maximum is where the function switches from increasing to decreasing, marked by a peak.
• A relative minimum is where the function transitions from decreasing to increasing, marked by a valley.

In the given exercise, however, since no critical points were found initially, we determine that there are no relative extrema in the domain of the function.

Critical Points

Critical points are where a function's first derivative is zero or undefined. These points are crucial because they are potential locations for relative extrema.

To find critical points:

• Take the derivative of the function.
• Solve for when this derivative equals zero or does not exist.

In the case of the function \(f(x) = \sqrt{x^2 + 1}\), the first derivative \(f'(x) = \frac{x}{\sqrt{x^2 + 1}}\) does not have real solutions when set to zero, and it is not undefined within the domain of the function.

This means that there are no critical points, indicating no changes that might suggest a maximum or minimum value within the function's range.

Concavity

Concavity describes the "curve" of a function's graph, showing whether it curves upwards or downwards. It is crucial for understanding the behavior and shape of the function.

A key way to determine concavity is to examine the second derivative:

• If the second derivative is positive, the graph is concave up, like a "U" shape.
• If the second derivative is negative, the graph is concave down, like an upside-down "U" shape.

For the function at hand, \(f''(x) = \frac{1}{(x^2 + 1)^{3/2}}\) is always positive for all real numbers because neither \(1\) nor \(x^2 + 1\) can be negative or zero. Therefore, the function is concave up everywhere, supporting the conclusion of no relative maxima or minima.

First Derivative

The first derivative of a function gives us the slope of the function's graph at any given point, indicating whether the function is increasing or decreasing at that point.

It helps in determining critical points where the slope is zero, which are potential spots for relative extrema.

• Positive first derivative: the function is increasing.
• Negative first derivative: the function is decreasing.
• First derivative equals zero: potential for a change in direction.

In the exercise, the first derivative \(f'(x) = \frac{x}{\sqrt{x^2 + 1}}\) describes the instantaneous rate of change of the function \(f(x) = \sqrt{x^2 + 1}\). Since it yields no zeros or discontinuities in the interval, it confirms the absence of critical points, crucial for assessing potential relative extrema.