Problem 12 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find equations for the inverses of the functions defined by (a) \(F_{1}(x)=\frac{1}{x+1}\) (b) \(\quad F_{2}(x)=\frac{x}{x+1}\) (c) \(\quad F_{3}(x)=1+2^{x}\) (d) \(\quad F_{4}(x)=\log _{2} x-\log _{2}(x+1)\) (e) \(\quad F_{5}(x)=10^{-x^{2}}\) for \(\mathrm{x} \geq 0\) (f) \(\quad F_{6}(z)=\frac{z+\frac{1}{z}}{2}\) for \(\mathrm{z} \geq 1\) (g) \(\quad F_{7}(x)=\frac{2^{x}-2^{-x}}{2}\)

Step-by-Step Solution

Answer

(a) \( F_1^{-1}(x) = \frac{1-x}{x} \). (b) \( F_2^{-1}(x) = \frac{-x}{x-1} \). (c) \( F_3^{-1}(x) = \log_2(x-1) \). (d) \( F_4^{-1}(x) = \log_2\left(\frac{x}{1-x}\right) \). (e) \( F_5^{-1}(x) = \sqrt{-\log_{10}(x)} \). (f) \( F_6^{-1}(z) \) involves solving quadratic. (g) \( F_7^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \).

Step 1: Solve for the inverse of \( F_1(x) = \frac{1}{x+1} \)

To find the inverse, set \( y = \frac{1}{x + 1} \). Swap \( x \) and \( y \) giving \( x = \frac{1}{y + 1} \). Solve for \( y \) by multiplying both sides by \( y + 1 \) to get \( x(y + 1) = 1 \). Distribute \( x \), yielding \( xy + x = 1 \). Isolate \( y \): \( xy = 1 - x \), then \( y = \frac{1 - x}{x} \). So, the inverse function is \( F_1^{-1}(x) = \frac{1-x}{x} \).

Step 2: Solve for the inverse of \( F_2(x) = \frac{x}{x+1} \)

Set \( y = \frac{x}{x+1} \), then swap to get \( x = \frac{y}{y+1} \). Multiply both sides by \( y+1 \), giving \( x(y+1) = y \). Distribute \( x \): \( xy + x = y \). Get terms involving \( y \) on one side: \( xy - y = -x \). Factor \( y \) out: \( y(x-1) = -x \). Solve for \( y \): \( y = \frac{-x}{x-1} \). Thus, \( F_2^{-1}(x) = \frac{-x}{x-1} \).

Step 3: Solve for the inverse of \( F_3(x) = 1 + 2^x \)

Set \( y = 1 + 2^x \), and swap \( x \) and \( y \) to get \( x = 1 + 2^y \). Subtract 1 from both sides: \( x - 1 = 2^y \). Take the logarithm base 2 of both sides: \( \log_2(x - 1) = y \). Hence, \( F_3^{-1}(x) = \log_2(x - 1) \).

Step 4: Solve for the inverse of \( F_4(x) = \log_2 x - \log_2(x+1) \)

Recognize \( F_4(x) \) as \( \log_2\left(\frac{x}{x+1}\right) \). Set \( y = \log_2\left(\frac{x}{x+1}\right) \), swap to have \( x = \frac{t}{t+1} \) with \( t = 2^y \). Solve for \( t \): \( x(t+1) = t \), or \( xt + x = t \), simplify to \( x = t(1-x) \), hence \( t = \frac{x}{1-x} \). Since \( t = 2^y \), so \( y = \log_2\left(\frac{x}{1-x}\right) \). Therefore, \( F_4^{-1}(x) = \log_2\left(\frac{x}{1-x}\right) \).

Step 5: Solve for the inverse of \( F_5(x) = 10^{-x^2} \) for \( x \geq 0 \)

Let \( y = 10^{-x^2} \). Set \( x = 10^{-y^2} \) and solve for \( x \). Take the log base 10 of both sides: \( \log_{10}(y) = -x^2 \). Divide by -1: \( x^2 = -\log_{10}(y) \). Assuming \( x \geq 0 \), \( x = \sqrt{-\log_{10}(y)} \). Hence, \( F_5^{-1}(x) = \sqrt{-\log_{10}(x)} \).

Step 6: Solve for the inverse of \( F_6(z) = \frac{z + \frac{1}{z}}{2} \) for \( z \geq 1 \)

Set \( y = \frac{z + \frac{1}{z}}{2} \), then swap to \( z = \frac{y + \frac{1}{y}}{2} \). Multiply both sides by 2 to simplify: \( 2z = y + \frac{1}{y} \). Arrange to get a quadratic form: \( 2zy = y^2 + 1 \). This can be solved using the quadratic formula for \( y \). Therefore, the inverse function requires simplification but general solution involves \( y = \frac{2z \pm \sqrt{(2z)^2 - 4}}{2} \) and checking which value makes sense \( y \geq 1 \).

Step 7: Solve for the inverse of \( F_7(x) = \frac{2^x - 2^{-x}}{2} \)

Recognize this as the hyperbolic sine function: \( F_7(x) = \sinh(x) \). Use the hyperbolic identity for inverse: \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \). Thus, \( F_7^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \).

Key Concepts

Function Manipulation

Function manipulation is a powerful mathematical skill that involves modifying and rearranging functions to find their inverse. To understand this, consider a function that maps input values (independent variable) to specific output values (dependent variable). The goal of finding an inverse function is to reverse this mapping process.
To manipulate a function effectively, follow these steps:

• Set the function equal to a variable, typically represented as \( y \).
• Swap the dependent and independent variables; change \( x \) to \( y \) and \( y \) to \( x \).
• Solve the equation for \( y \) to express \( y \) exclusively in terms of \( x \).
• Re-label \( y \) as the inverse function of the original, often written as \( f^{-1}(x) \).

Remember, not all functions have inverses unless they pass both the horizontal line test (for injectivity) and the vertical line test (for functionality). Proper manipulation often leads to a simplified inverse expression, revealing insights into the behavior of the original function.

Logarithms

Logarithms are mathematical tools used to find the exponent needed for a given base to achieve a certain number. They are inverse operations of exponentiation, playing a crucial role in solving exponential equations and manipulating functions.
For example, when needing to solve for the inverse of an exponential function, such as \( y = 2^x \), using logarithms is essential. To find the inverse, you would swap the variables and take the logarithm of both sides:

• Start with \( x = 2^y \).
• Take the logarithm base 2: \( \, \log_2(x) = y \).

This shows that the inverse of \( y = 2^x \) is \( y = \log_2(x) \). Logarithms transform multiplicative relationships into additive ones, making complex calculations more manageable. Moreover, the properties like \( \log_a(bc) = \log_a(b) + \log_a(c) \) and \( \log_a(b^c) = c\log_a(b) \) help simplify expressions during function manipulation.

Exponential Functions

Exponential functions are crucial in representing growth and decay processes in mathematics. They are characterized by a constant base raised to the power of a variable, typically noted as \( y = a^x \), where \( a \) is a positive constant.

• Exponential growth: When the base \( a > 1 \), the function represents growth, rapidly increasing its value as \( x \) increases.
• Exponential decay: For \( 0 < a < 1 \), the function describes decay, decreasing rapidly as \( x \) increases.

Finding the inverse of an exponential function involves using logarithms, as these functions are composed of repeated multiplications. For instance, solving \( y = 2^x \) for its inverse involves swapping the variables and taking the logarithm: \( x = 2^y \) then yields \( y = \log_2(x) \), demonstrating that the function mirroring the behavior of the exponential curve is itself defined by a logarithmic function.
Exponential functions are prevalent across various disciplines, prominently featured in population growth models, radioactive decay, and interest calculations in finance.

Algebraic Equations

Algebraic equations involve mathematical statements where two expressions are set equal to each other, often containing variables, constants, and operators. Solving them requires a systematic approach to isolate variables and simplify expressions.
When working with inverse functions, algebraic techniques come into play, especially when manipulating the original function to solve for the dependent variable \( y \). For example, consider the manipulation of \( y = \frac{x}{x+1} \) to find its inverse:

• Swap the variables: \( x = \frac{y}{y+1} \).
• Clear the fraction by multiplying both sides by \( y+1 \): \( x(y+1) = y \).
• Expand and rearrange: \( xy + x = y \).
• Isolate terms with \( y \) on one side: \( xy - y = -x \).
• Solve for \( y \) by factoring: \( y = \frac{-x}{x-1} \).

This procedure highlights algebra’s role in transforming complex equations into simplified forms, enabling us to ascertain solutions that offer insights into the behavior of inverse relationships.