Question

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

Short answer

The function \(f(x)=\sqrt{x^{2}+1}\) has one relative minimum at \(x=0\). There are no relative maximum points.

Step-by-step solution

Differentiate the function

First differentiate the function with respect to \(x\). Using the chain rule, the derivative of \(f(x)=\sqrt{x^{2}+1}\) is \(f'(x) = \frac{x}{\sqrt{x^{2}+1}}\).

Calculate the Critical Points

Critical points occur where the derivative of the function is equal to zero or undefined. The derivative \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\) is defined for all real numbers and equating it to zero gives \(x=0\). So there is only one critical point which is at \(x=0\).

Compute the second derivative

The second derivative of the function can be computed as \(f''(x)= -\frac{x^2}{(x^2+1)^(3/2)} + \frac{1}{(x^2+1)^(3/2)} = \frac{1 - x^2}{(x^2+1)^(3/2)}\).

Apply the Second-Derivative Test

In Second-Derivative Test, if \(f''(x) > 0\), then \(x\) is a relative minimum. If \(f''(x) < 0\), then \(x\) is a relative maximum. If \(f''(x) = 0\), then the test is inconclusive. Evaluate \(f''(x)\) at \(x=0\), we find \(f''(0) = \frac{1 - 0^2}{(0^2+1)^(3/2)} = 1 > 0 \). Thus, the function \(f(x)=\sqrt{x^{2}+1}\) has a relative minimum at \(x=0\).

Key concepts

Relative Extrema

Relative extrema are points in a function where the function reaches a local maximum or minimum value. These are points where you can see peaks (maximum) or valleys (minimum) when looking at the graph of the function. Relative maximum is a point

• where the function changes direction from increasing to decreasing.
• Imagine the top of a hill.

Similarly, a relative minimum is a point

• where the function changes direction from decreasing to increasing.
• Picture the bottom of a valley.

In this exercise, we focused on finding where these points occur by using the powerful Second-Derivative Test, which helps confirm if a critical point is indeed a relative extremum.

Critical Points

Critical points are values of the variable where the derivative of a function is zero or undefined. A critical point is a potential candidate for where the function has relative extrema. To find critical points:

• First, find the derivative of the function.
• Second, set the derivative equal to zero to solve for the variable.
• Also, check where the derivative might be undefined.

In the function given, the expression for the derivative is well-defined for all real numbers. When equating the derivative to zero, we found a critical point at x = 0.

Derivative

The derivative of a function is a fundamental concept in calculus that measures how the function's value changes as its input changes. Essentially, the derivative gives us the slope of the tangent line to the function at any point. Calculating a derivative involves rules like:

• The power rule for basic polynomials.
• The chain rule for composite functions.

The derivative is crucial for identifying critical points, enabling us to understand the behavior and timing of changes in a function. For the exercise's function, we used the chain rule to find the derivative as \(f'(x) = \frac{x}{\sqrt{x^{2}+1}}\).

Chain Rule

The chain rule is a method used in calculus for differentiating composite functions. When functions are nested within each other, the chain rule helps us take their derivatives efficiently. The chain rule states that to take the derivative of a composite function, differentiate the outer function while multiplying by the derivative of the inner function. Steps in the chain rule:

• Identify the outer and inner functions.
• Calculate the derivative of the outer function.
• Multiply that result by the derivative of the inner function.

In the exercise, we had \(f(x) = \sqrt{x^2 + 1}\) where the outer function is the square root, and the inner function is \(x^2 + 1\). Using the chain rule, we found \(f'(x) = \frac{x}{\sqrt{x^{2}+1}}\), which is key to identifying critical points.