Problem 19 - Calculus for Biology and Medicine

Question

Let $$f(x)=\frac{x^{2}}{1+x^{2}}, x \in \mathbf{R}$$ (a) Determine where \(f(x)\) is increasing and where it is decreasing. (b) Where is the function concave up and where is it concave down? Find all inflection points of \(f(x)\). (c) Find \(\lim _{x \rightarrow \pm \infty} f(x)\) and decide whether \(f(x)\) has a horizontal asymptote. (d) Sketch the graph of \(f(x)\) together with its asymptotes and inflection points (if they exist).

Step-by-Step Solution

Answer

\(f(x)\) is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\); concave up on \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), down elsewhere; \(x = \pm \frac{1}{\sqrt{3}}\) are inflection points; \(y=1\) is a horizontal asymptote.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first need the first derivative. For \( f(x) = \frac{x^2}{1+x^2} \), use the quotient rule: \( f'(x) = \frac{(2x)(1+x^2) - x^2(2x)}{(1+x^2)^2} = \frac{2x}{(1+x^2)^2} \).

Step 2: Solve f'(x) = 0 to Find Critical Points

Set the first derivative equal to zero: \( \frac{2x}{(1+x^2)^2} = 0 \). The numerator, \(2x\), dictates the zeros, so \(x = 0\) is the critical point. This divides the number line into intervals for testing.

Step 3: Test for Increasing and Decreasing Intervals

Test the sign of \( f'(x) \) on intervals \((-\infty, 0)\) and \((0, \infty)\). Choose test points: \( f'(-1) < 0 \) and \( f'(1) > 0 \). Thus, \( f(x) \) is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).

Step 4: Find the Second Derivative

To determine concavity, find the second derivative. Differentiate \( f'(x) = \frac{2x}{(1+x^2)^2} \) using the quotient rule again: \( f''(x) = \frac{2(1+x^2)^2 - 8x^2(1+x^2)}{(1+x^2)^4} = \frac{2 - 6x^2}{(1+x^2)^3} \).

Step 5: Solve f''(x) = 0 to Find Inflection Points

Set the second derivative equal to zero: \( \frac{2 - 6x^2}{(1+x^2)^3} = 0 \). Solving gives \(x^2 = \frac{1}{3}\), so \(x = \pm \frac{1}{\sqrt{3}}\) are potential inflection points.

Step 6: Test Concavity on Intervals

Test the sign of \( f''(x) \) in the intervals \((-\infty, -\frac{1}{\sqrt{3}})\), \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), and \((\frac{1}{\sqrt{3}}, \infty)\). On \((-\infty, -\frac{1}{\sqrt{3}})\) and \((\frac{1}{\sqrt{3}}, \infty)\), \(f''(x) < 0\), and on \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), \(f''(x) > 0\). Thus, the function is concave down on \((-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty)\) and concave up on \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\). The inflection points are at \(x = \pm \frac{1}{\sqrt{3}}\).

Step 7: Evaluate Limits for Horizontal Asymptotes

Calculate \( \lim_{x \to \pm \infty} \frac{x^2}{1+x^2} \). This limit evaluates to 1 by dividing numerator and denominator by \(x^2\). Therefore, \( y = 1 \) is a horizontal asymptote.

Step 8: Sketch the Graph of f(x)

The graph should include the points of increase and decrease, concavity, the horizontal asymptote at \(y=1\), and the inflection points at \(x=\pm \frac{1}{\sqrt{3}}\). The graph decreases on \((-\infty, 0)\), increases on \((0, \infty)\), is concave up between \(-\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\), and concave down elsewhere.

Key Concepts

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing is key to analyzing its behavior. For the function given, \( f(x) = \frac{x^2}{1+x^2} \), we can find these intervals by looking at the first derivative, \( f'(x) \). The first derivative tells us the slope of the tangent line at any point on the curve. If \( f'(x) > 0 \), the function is increasing, and if \( f'(x) < 0 \), it is decreasing.

To find \( f'(x) \), we use the quotient rule. The result is \( f'(x) = \frac{2x}{(1+x^2)^2} \). We then find where \( f'(x) = 0 \), which gives critical points where the slope of the tangent is zero, potentially marking a point where the function shifts from increasing to decreasing or vice versa.

Here, \( x = 0 \) is the only critical point. Testing intervals around this point: on \( (-\infty, 0) \), \( f'(x) < 0 \), indicating the function is decreasing; on \( (0, \infty) \), \( f'(x) > 0 \), which shows the function is increasing.

Concavity and Inflection Points

Concavity tells us how the curves bend. If the curve is concave up, like a cup, it opens upwards. If it bends down, like a frown, it's concave down. To determine concavity, we look at the second derivative, \( f''(x) \). If \( f''(x) > 0 \), the curve is concave up; if \( f''(x) < 0 \), it's concave down.

Calculating \( f''(x) \) for \( f(x) = \frac{x^2}{1+x^2} \), we find \( f''(x) = \frac{2 - 6x^2}{(1+x^2)^3} \). Setting \( f''(x) = 0 \) to find inflection points, we solve for \( x \) and find \( x = \pm \frac{1}{\sqrt{3}} \). Inflection points are where the function changes concavity.

By testing the intervals around our inflection points, we learn: for \( x \in (-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty) \), the function is concave down as \( f''(x) < 0 \); for \( x \in (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \), it's concave up as \( f''(x) > 0 \). Thus, there is a change in concavity at \( x = \pm \frac{1}{\sqrt{3}} \).

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \( x \) moves towards positive or negative infinity. It shows a line that the graph approaches but never touches. This is particularly useful in functions that flatten out.

To find horizontal asymptotes for our function \( f(x) = \frac{x^2}{1+x^2} \), we calculate \( \lim_{x \to \pm \infty} \frac{x^2}{1+x^2} \). By simplifying, we divide both the numerator and the denominator by \( x^2 \), which leads to the limits equal to 1 as \( x \) approaches both positive and negative infinity.

Thus, the horizontal asymptote is \( y = 1 \). This means no matter how large or small \( x \) gets, \( f(x) \) will approach the line \( y = 1 \) from below as \( x \) increases, offering a flat, leveling behavior.