Problem 34 - Calculus for Biology and Medicine

Question

Determine whether the functions have absolute maxima and minima, and, if so, find their coordinates. Find inflection points. Find the intervals on which the function is increasing, on which it is decreasing, on which it is concave up, and on which it is concave down. Sketch the graph of each function. $$ y=\ln \left(x^{2}+1\right), x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function has an absolute minimum at (0,0) and inflection points at (-1, \ln(2)) and (1, \ln(2)). It increases on (0,\(\infty\)), decreases on (\(-\infty\),0), is concave up on (-1,1) and concave down elsewhere.

Step 1: Identify the Domain

The function is given by \( y = \ln(x^2 + 1) \). Since \( \ln(x^2 + 1) \) is defined for all real numbers because \( x^2 + 1 > 0 \) for any real \( x \), the domain is \( x \in \mathbb{R} \).

Step 2: Find First Derivative

Calculate the first derivative of the function to find critical points. The derivative \( y' \) is computed as:\[y' = \frac{d}{dx}\left( \ln(x^2 + 1) \right) = \frac{2x}{x^2 + 1}\]

Step 3: Determine Critical Points

Set the first derivative equal to zero to find critical points:\[\frac{2x}{x^2 + 1} = 0\]Since \( x^2 + 1 \) is never zero, the equation simplifies to \( 2x = 0 \) which gives \( x = 0 \) as the only critical point.

Step 4: First Derivative Test

Use the first derivative test to check whether the critical point \( x = 0 \) is a maximum, minimum, or neither. Evaluate \( y' \) near \( x = 0 \):- For \( x < 0 \), \( y' = \frac{2x}{x^2 + 1} < 0 \), so the function is decreasing.- For \( x > 0 \), \( y' = \frac{2x}{x^2 + 1} > 0 \), so the function is increasing.Therefore, \( x = 0 \) is a local minimum.

Step 5: Identify Function Behavior at Critical Point

Evaluate the function at \( x = 0 \):\[y(0) = \ln(0^2 + 1) = \ln(1) = 0\]The function has a local minimum at \( (0, 0) \). Since \( \ln(x^2 + 1) \) is always positive for \( x eq 0 \), this is also an absolute minimum.

Step 6: Find Second Derivative

The second derivative \( y'' \) helps in determining the concavity of the function:\[y'' = \frac{d}{dx}\left( \frac{2x}{x^2 + 1} \right) = \frac{2(x^2 + 1) - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}\]

Step 7: Determine Inflection Points

Set the second derivative \( y'' \) to zero to find inflection points:\[\frac{2 - 2x^2}{(x^2 + 1)^2} = 0 \Rightarrow 2 - 2x^2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\]These are the inflection points.

Step 8: Second Derivative Test

Evaluate the concavity by testing the intervals determined by inflection points:- For \( x < -1 \) and \( x > 1 \), \( y'' < 0 \), indicating the function is concave down.- For \( -1 < x < 1 \), \( y'' > 0 \), indicating the function is concave up.

Step 9: Summarize Intervals and Sketch

The function is:- Increasing on \( (0, \infty) \) and decreasing on \( (-\infty, 0) \).- Concave up on \( (-1, 1) \) and concave down on \( (-\infty, -1) \cup (1, \infty) \).- Absolute and local minimum at \( (0, 0) \).- Inflection points at \( (-1, \ln(2)) \) and \( (1, \ln(2)) \).Based on these points, sketch the approximate graph shape.

Key Concepts

Inflection Points

In calculus, inflection points are where a function changes its concavity from concave up to concave down or vice versa. These points can indicate a change in the direction of the curvature of the graph. To find inflection points, we must calculate the second derivative of the function and set it equal to zero.
For the function given in the exercise, \[ y = \ln(x^2 + 1) \] the second derivative is \[ y'' = \frac{2 - 2x^2}{(x^2 + 1)^2} \]. Setting this equal to zero helps find the inflection points:

• \(2 - 2x^2 = 0\)
• \(2x^2 = 2\)
• \(x^2 = 1\)
• \(x = \pm 1\)

So the inflection points are at \(x = -1\) and \(x = 1\). These are the coordinates where the curvature of the graph of the function changes.

Concavity

Understanding concavity helps identify the nature of a function's curvature. A function is concave up when it appears like a cup opening upwards, and concave down when it appears like a cup opening downwards. This can be determined using the sign of the second derivative:

• If \( y'' > 0 \), the function is concave up.
• If \( y'' < 0 \), the function is concave down.

For the given function, the intervals of concavity are assessed by analyzing sign changes in \[ y'' = \frac{2 - 2x^2}{(x^2 + 1)^2} \].

• For \( x < -1 \) and \( x > 1 \), \( y'' < 0 \), so the function is concave down on \( (-\infty, -1) \cup (1, \infty) \).
• For \( -1 < x < 1 \), \( y'' > 0 \), so the function is concave up on \( (-1, 1) \).

These concavity intervals help us understand where the graph bends upwards or downwards.

Derivative Test

The derivative test is a useful tool for determining the critical points of a function. By analyzing the sign of the first derivative, we can describe where the function is increasing or decreasing:

• If the first derivative \( y' > 0 \), the function is increasing.
• If the first derivative \( y' < 0 \), the function is decreasing.

For the function \[ y = \ln(x^2 + 1) \], the first derivative is \[ y' = \frac{2x}{x^2 + 1} \]. By setting \( y' = 0 \), we find the critical point at \( x = 0 \).
Using the first derivative test around \( x = 0 \):

• For \( x < 0 \), \( y' < 0 \), indicating decreasing.
• For \( x > 0 \), \( y' > 0 \), indicating increasing.

This analysis suggests that \( x = 0 \) is a local minimum point, as the function transitions from decreasing to increasing.

Domain of a Function

The domain of a function includes all the input values for which the function is well-defined or makes sense. For logarithmic functions, the domain comprises values for which the argument is positive, since the logarithm of a non-positive number is undefined.
With the function \[ y = \ln(x^2 + 1) \], the expression \( x^2 + 1 \) is always positive for all real values of \( x \), as a square plus one is never zero or negative.
Therefore, the domain of this function is the entire set of real numbers, \( x \in \mathbb{R} \). This means you can substitute any real number for \( x \) in this function, without it resulting in an undefined mathematical operation. Knowing the domain ensures what inputs are valid for the function.