Problem 39 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\frac{x}{x+1}, \quad x>-1 $$ (a) Use a graphing calculator to graph \(f(x)\). (b) On the basis of the graph in (a), determine the range of \(f(x)\). (c) For which values of \(x\) is \(f(x)=3 / 4\) ? (d) On the basis of the graph in (a), determine how many solutions \(f(x)=a\) has, where \(a\) is in the range of \(f(x)\).

Step-by-Step Solution

Answer

(b) Range is \((-1, 1)\); (c) \( x = 3 \); (d) One solution for each \( a \) in \((-1, 1)\).

Step 1: Graphical Visualization

Using a graphing calculator, plot the function \( f(x) = \frac{x}{x+1} \) for \( x > -1 \). Notice that the function approaches \( y = 1 \) as \( x \to \infty \) and approaches \( y = -1 \) as \( x \to -1^+ \). It also crosses the x-axis at the origin as \( f(0) = 0 \).

Step 2: Analyzing Range from Graph

From the plotted graph, observe that \( f(x) \) appears to cover all y-values between \(-1\) and \(1\), but it does not actually reach \(1\), as it asymptotically approaches \(1\). Therefore, the range of \( f(x) \) is \( (-1, 1) \).

Step 3: Finding Values for Specific Output

To find \( x \) such that \( f(x) = \frac{3}{4} \), set up the equation \( \frac{x}{x+1} = \frac{3}{4} \). Cross-multiplying gives \( 4x = 3(x + 1) \). Solving for \( x \), we distribute the 3, getting \( 4x = 3x + 3 \). Subtract \( 3x \) from both sides to obtain \( x = 3 \).

Step 4: Solution Verification

Verify the solution for \( f(x) = \frac{3}{4} \) by substituting \( x = 3 \) back into the function: \( f(3) = \frac{3}{3+1} = \frac{3}{4} \). This confirms that \( x = 3 \) is indeed a solution.

Step 5: Determining Number of Solutions

From the graph in (a), notice that the function \( f(x) \) is continuous and strictly increasing over its domain \( x > -1 \). Therefore, for any \( a \) in the range \((-1, 1)\), there is exactly one value of \( x \) that will make \( f(x) = a \). Thus, the number of solutions is one for each \( a \) in the range.

Key Concepts

Understanding the Range of Functions

The "range" in a function is a set of possible output values (certain y-values) that a function can produce. Consider an equation where the variable `x` is in the domain; the range will include all resultant `y-values`. In the given problem with the function \( f(x) = \frac{x}{x+1} \), the range is all the values that \( f(x) \) can achieve. Based on the graph and behavior of the function, we identify

• The function approaches 1 but never actually reaches it.
• The function also gets close to -1 as \( x \) moves towards -1 from the right.

Thus, the range is all real numbers between -1 and 1, i.e., \((-1, 1)\). It's essential to understand that the function never achieves the endpoints of the range due to its asymptotic nature at those boundaries.

Graphical Analysis Basics

Graphical analysis involves visualizing functions to understand their behavior, and often provides intuition on the range, intercepts, and asymptotic behavior. When you graph \( f(x) = \frac{x}{x+1} \) using a calculator or plotting software:

• You'll observe the characteristic curves that indicate how the function behaves as \( x \) becomes very large (approaching y = 1) and very small (approaching y = -1).
• Determining the function's intersections with the axes, the x-intercept occurs where \( f(0) = 0 \).

Graphical analysis provides a visual representation of these concepts, making it easier to see the function's overall trend and more subtle patterns.

Asymptotic Behavior in Functions

Asymptotic behavior describes how functions behave as inputs grow very large or very close to certain points. In calculus, understanding asymptotic behavior helps predict long-term trends or behavior near particular boundaries. For our function \( f(x) = \frac{x}{x+1} \):

• The function approaches but never reaches y = 1 as the value of \( x \) increases towards infinity. This creates a horizontal asymptote at y = 1.
• As \( x \) approaches -1 from the right, the function tends towards negative infinity. This indicates a vertical asymptote near x = -1.

Recognizing these behaviors is useful when sketching graphs and analyzing the function's limits. Asymptotes are crucial in describing how the function stabilizes or diverges around these critical areas.

Solving Equations Algebraically

Solving equations algebraically involves finding the exact value of variables that satisfy a given equation. For example, to find where \( f(x) = \frac{3}{4} \), we set the function equal to \( \frac{3}{4} \):

• The equation to solve is \( \frac{x}{x+1} = \frac{3}{4} \).
• By cross-multiplying, we get \( 4x = 3(x + 1) \).
• Solving this, distribute to obtain \( 4x = 3x + 3 \), then subtract \( 3x \) resulting in \( x = 3 \).

Checking the solution by substituting back into the original function verifies that \( x = 3 \) is a valid solution, as \( f(3) = \frac{3}{4} \). This equation-solving method is fundamental for finding precise solutions analytically.