Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(f(x)=x^{2}+x\) and \(g(x)=2 /(x+3)\), find each value. (a) \((f-g)(2)\) (b) \((f / g)(1)\) (c) \(g^{2}(3)\) (d) \((f \circ g)(1)\) (e) \((g \circ f)(1)\) (f) \((g \circ g)(3)\)

Step-by-Step Solution

Answer

(a) \(\frac{28}{5}\), (b) 4, (c) \(\frac{1}{9}\), (d) \(\frac{3}{4}\), (e) \(\frac{2}{5}\), (f) \(\frac{3}{5}\).

Step 1: Calculate \(f(x)\) and \(g(x)\)

We start by calculating the values of \(f(x)\) and \(g(x)\) using their respective equations.- For \(f(x) = x^2 + x\), substituting the value of \(x\) will give us the function value.- For \(g(x) = \frac{2}{x+3}\), substituting the value of \(x\) will give us the function value.

Step 2: Compute \((f-g)(2)\)

First calculate \(f(2)\):\[ f(2) = 2^2 + 2 = 4 + 2 = 6 \]Then calculate \(g(2)\):\[ g(2) = \frac{2}{2+3} = \frac{2}{5} \]Now subtract: \[ (f-g)(2) = f(2) - g(2) = 6 - \frac{2}{5} = \frac{30}{5} - \frac{2}{5} = \frac{28}{5} \]

Step 3: Compute \((f / g)(1)\)

Calculate \(f(1)\):\[ f(1) = 1^2 + 1 = 1 + 1 = 2 \]Then calculate \(g(1)\):\[ g(1) = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2} \]Now divide:\[ (f / g)(1) = \frac{f(1)}{g(1)} = \frac{2}{\frac{1}{2}} = 2 \times 2 = 4 \]

Step 4: Compute \(g^2(3)\)

Calculate \(g(3)\):\[ g(3) = \frac{2}{3+3} = \frac{2}{6} = \frac{1}{3} \]Then square the result:\[ g^2(3) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

Step 5: Compute \((f \circ g)(1)\)

First calculate \(g(1)\) (already done in Step 3):\[ g(1) = \frac{1}{2} \]Then plug this into \(f(x)\):\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]

Step 6: Compute \((g \circ f)(1)\)

First calculate \(f(1)\) (done in Step 3):\[ f(1) = 2 \]Now plug this into \(g(x)\):\[ g(2) = \frac{2}{2 + 3} = \frac{2}{5} \]

Step 7: Compute \((g \circ g)(3)\)

First calculate \(g(3)\) (from Step 4):\[ g(3) = \frac{1}{3} \]Then plug this into \(g(x)\):\[ g\left(\frac{1}{3}\right) = \frac{2}{\frac{1}{3} + 3} = \frac{2}{\frac{1}{3} + \frac{9}{3}} = \frac{2}{\frac{10}{3}} = \frac{2 \times 3}{10} = \frac{6}{10} = \frac{3}{5} \]

Key Concepts

Composite Functions

Imagine having two functions and plugging one into the other. That's what composite functions are all about. When we write \(f \circ g\), it means substituting the entire function of \(g(x)\) into each \(x\) in \(f(x)\).
For example, if \(f(x) = x^{2} + x\) and \(g(x) = \frac{2}{x+3}\), the composite function \(f \circ g(1)\) becomes inserting \(g(1)\) into \(f(x)\), and evaluating:- Firstly, find \(g(1)\): \[ g(1) = \frac{2}{1 + 3} = \frac{1}{2} \]- Next, substitute this value into \(f\): \[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]Composite functions allow us to create new behaviors and dependencies between two or more initial functions. This is essential for complex calculations where combining functionality from different mathematical rules is necessary.

Function Subtraction

Function subtraction involves subtracting the value of one function from another at the same input value. This is denoted as \(f-g\). Consider two functions, such as \(f(x) = x^{2} + x\) and \(g(x) = \frac{2}{x+3}\). To find \(f-g(2)\), follow these steps:

• Evaluate \(f(2)\): \[ f(2) = 2^2 + 2 = 6 \]
• Evaluate \(g(2)\): \[ g(2) = \frac{2}{2+3} = \frac{2}{5} \]
• Subtract the two values: \[ (f-g)(2) = 6 - \frac{2}{5} = \frac{28}{5} \]

Function subtraction helps in comparing outputs and understanding how two functions differ in magnitude at any given input. It serves as a foundational operation in function analysis.

Function Division

Function division entails dividing the output of one function by another for a specific input. Symbolically represented as \(\frac{f}{g}\), it is crucial to ensure that the divisor \(g(x)\) is not zero to avoid undefined operations.
For instance, given \(f(x) = x^{2} + x\) and \(g(x) = \frac{2}{x+3}\), to calculate \(\frac{f}{g}(1)\):

• Find \(f(1)\): \[ f(1) = 1^2 + 1 = 2 \]
• Find \(g(1)\): \[ g(1) = \frac{1}{2} \]
• Perform the division: \[ \frac{f(1)}{g(1)} = \frac{2}{\frac{1}{2}} = 4 \]

Function division forms the basis for understanding how one function behaves relative to another under division,enabling variable manipulation and dynamic system interpretations.

Function Squaring

Function squaring involves taking the output of a function and squaring it.It is represented as \(g^2(x)\) or \((g(x))^2\).
The process is straightforward:

• Compute the function value at a specific point, for example, \(g(3)\): \[ g(3) = \frac{1}{3} \]
• Square the result to get \[ g^2(3) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

This operation with squaring emphasizes the growth or decrease of a function input’s effect squared,impacting range and variance, and is frequently encountered in probability and statistics to determine variance or standard deviations.