Problem 60 - Calculus Early Transcendentals

Question

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=\frac{1}{1+x^{2}}$$

Step-by-Step Solution

Answer

Question: Determine the intervals on which the function $$f(x) = \frac{1}{1 + x^2}$$ is concave up and concave down, and identify the inflection points. Answer: The function is concave up on the intervals $$(-\infty, -\frac{1}{\sqrt{3}})$$ and $$(\frac{1}{\sqrt{3}}, \infty)$$, concave down on the interval $$(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$, and has inflection points at $$\left(-\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$ and $$\left(\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$.

Step 1: Find the first derivative of $$f(x)$$.

To begin, we need to find the first derivative of the given function $$f(x) = \frac{1}{1 + x^2}$$. We will use the power rule and the chain rule to do this: $$f'(x) = -\frac{2x}{(1+x^2)^2}$$

Step 2: Find the second derivative of $$f(x)$$.

Now, we need to find the second derivative of $$f(x)$$. We will again use the power rule and the chain rule for differentiation: $$f''(x) = \frac{2(3x^2 - 1)}{(1+x^2)^3}$$

Step 3: Determine the critical points.

Critical points occur when the second derivative of a function is equal to 0 or undefined. Our second derivative $$f''(x)$$ is never undefined, so we only need to solve for $$x$$ when $$f''(x) = 0$$: $$\frac{2(3x^2 - 1)}{(1+x^2)^3} = 0$$ The numerator must be 0, so we have: $$3x^2 - 1 = 0$$ $$x^2 = \frac{1}{3}$$ $$x = \pm\frac{1}{\sqrt{3}}$$ Thus, our critical points are $$x = -\frac{1}{\sqrt{3}}$$ and $$x = \frac{1}{\sqrt{3}}$$.

Step 4: Test the intervals between the critical points.

We will now test the second derivative for an $$x$$-value in each of the intervals between the critical points to determine whether the function is concave up or concave down in that interval: 1. Interval $$(-\infty, -\frac{1}{\sqrt{3}})$$: If we choose $$x = -2$$, then $$f''(-2) > 0$$. Therefore, the function is concave up in this interval. 2. Interval $$(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$: If we choose $$x = 0$$, then $$f''(0) < 0$$. Therefore, the function is concave down in this interval. 3. Interval $$(\frac{1}{\sqrt{3}}, \infty)$$: If we choose $$x = 2$$, then $$f''(2) > 0$$. Therefore, the function is concave up in this interval.

Step 5: Identify the inflection points.

Inflection points occur where the concavity changes. From our intervals, we established that there are two such points at $$x = -\frac{1}{\sqrt{3}}$$ and $$x = \frac{1}{\sqrt{3}}$$, where the function changes from being concave up to concave down and vice versa. To find the corresponding $$y$$-values, we plug these $$x$$-values back into the original function: $$f\left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$$ $$f\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$$ So the inflection points are: $$\left(-\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$ and $$\left(\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$. In conclusion, - The function is concave up on the intervals $$(-\infty, -\frac{1}{\sqrt{3}})$$ and $$(\frac{1}{\sqrt{3}}, \infty)$$. - The function is concave down on the interval $$(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$. - The inflection points are $$\left(-\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$ and $$\left(\frac{1}{\sqrt{3}}, \frac{3}{4}\right)$$.

Key Concepts

First Derivative

The first derivative of a function, denoted as \(f'(x)\), provides essential information about the behavior of the function. In the context of concavity and inflection points, the first derivative tells us the slope of the tangent line at any given point on the curve. It helps us identify where the function is increasing or decreasing. For the function \(f(x) = \frac{1}{1 + x^2}\), the first derivative is calculated using differentiation rules like the power rule and chain rule. The result is \(f'(x) = -\frac{2x}{(1+x^2)^2}\).

This expression shows how the slope of the function changes depending on the value of \(x\). The negative sign indicates that the function starts decreasing as \(x\) moves away from 0 in either direction. While primarily used for finding extreme points (maximums and minimums), the first derivative also sets the stage for deeper analysis through the second derivative.

Second Derivative

The second derivative, \(f''(x)\), is a crucial tool for understanding the concavity of a function. It tells us how the rate of change is changing, which ultimately informs us whether the function is bending upwards or downwards. For the function \(f(x) = \frac{1}{1 + x^2}\), the second derivative is \(f''(x) = \frac{2(3x^2 - 1)}{(1+x^2)^3}\).

By setting \(f''(x) = 0\), we find potential inflection points, where the concavity might change. In our example, the expression simplifies into \(3x^2 - 1 = 0\), giving two solutions for \(x\): \(x = \pm\frac{1}{\sqrt{3}}\).

The second derivative tells us more than just where the function may switch concavity; it also helps pinpoint the nature of concavity:

• If \(f''(x) > 0\), the function is concave up (shaped like a bowl).
• If \(f''(x) < 0\), the function is concave down (shaped like a dome).

Analyzing these intervals where \(f''(x)\) changes sign helps us map the curve more accurately.

Inflection Points

An inflection point is where the curve changes its concavity from up to down or vice versa. It signifies a change in the 'bending' direction of the function. For \(f(x) = \frac{1}{1 + x^2}\), we discovered inflection points at \(x = \pm\frac{1}{\sqrt{3}}\), as these values cause the second derivative to equal zero, marking a potential change in concavity.

To confirm these are inflection points, we examined the intervals around \(x = -\frac{1}{\sqrt{3}}\) and \(x = \frac{1}{\sqrt{3}}\). By testing some \(x\)-values in these intervals, the function showed concavity changes:

• Concave up when \(x < -\frac{1}{\sqrt{3}}\) and \(x > \frac{1}{\sqrt{3}}\).
• Concave down between \(x = -\frac{1}{\sqrt{3}}\) and \(x = \frac{1}{\sqrt{3}}\).

The function transitions from being concave up to concave down at \(x = -\frac{1}{\sqrt{3}}\), and back to concave up at \(x = \frac{1}{\sqrt{3}}\). These points are crucial for understanding the overall graph of the function.