Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(f(x)=x+3\) and \(g(x)=x^{2}\), find each value. (a) \((f+g)(2)\) (b) \((f \cdot g)(0)\) (c) \((g / f)(3)\) (d) \((f \circ g)(1)\) (e) \((g \circ f)(1)\) (f) \((g \circ f)(-8)\)

Step-by-Step Solution

Answer

(a) 9, (b) 0, (c) \(\frac{3}{2}\), (d) 4, (e) 16, (f) 25.

Step 1: Finding \((f+g)(2)\)

First, we need to find the value of \(f(x)\) and \(g(x)\) at \(x = 2\). \(f(x) = x+3\), so \(f(2) = 2 + 3 = 5\). \(g(x) = x^2\), so \(g(2) = 2^2 = 4\). Therefore, \((f+g)(2) = f(2) + g(2) = 5 + 4 = 9\).

Step 2: Finding \((f \cdot g)(0)\)

Next, we find \(f(0)\) and \(g(0)\). \(f(0) = 0 + 3 = 3\) and \(g(0) = 0^2 = 0\). Therefore, \((f \cdot g)(0) = f(0) \cdot g(0) = 3 \cdot 0 = 0\).

Step 3: Finding \((g / f)(3)\)

We need \(f(3)\) and \(g(3)\). \(f(3) = 3 + 3 = 6\) and \(g(3) = 3^2 = 9\). So, \((g / f)(3) = \frac{g(3)}{f(3)} = \frac{9}{6} = \frac{3}{2}\).

Step 4: Finding \((f \circ g)(1)\)

This requires finding \(g(1)\) first, then substituting into \(f(x)\). \(g(1) = 1^2 = 1\). Then \(f(g(1)) = f(1) = 1 + 3 = 4\). So, \((f \circ g)(1) = 4\).

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

Find \(f(1)\) first, then substitute into \(g(x)\). \(f(1) = 1 + 3 = 4\). Then \(g(f(1)) = g(4) = 4^2 = 16\). So, \((g \circ f)(1) = 16\).

Step 6: Finding \((g \circ f)(-8)\)

First, calculate \(f(-8)\) then substitute into \(g(x)\). \(f(-8) = -8 + 3 = -5\). Then \(g(f(-8)) = g(-5) = (-5)^2 = 25\). So, \((g \circ f)(-8) = 25\).

Key Concepts

Composite Functions

Composite functions, often denoted by \(f \circ g\), involve creating a new function by combining two existing functions. It's like feeding the output of one function into another. For example, if you have functions \(f(x) = x + 3\) and \(g(x) = x^2\), a composite function \(f \circ g\) means you first apply \(g(x)\) and then use the result as the input for \(f(x)\).
This is called the composition of functions and is sometimes read as "f composed with g." Performing this operation may seem tricky at first, but it follows a consistent pattern:

• Compute the innermost function first.
• Use its output as the input to the outer function.

Understanding composite functions helps in breaking down complex calculations into manageable steps, making them a valuable tool in mathematics.

Function Composition

Function composition is a method where two functions are combined so that the output of one function becomes the input for the next. It's represented as \(f \circ g(x)\), translating to applying \(g(x)\) first and then applying \(f(x)\) to that result. Using the example functions \(f(x) = x + 3\) and \(g(x) = x^2\), to find \(f \circ g(1)\), you would:

• Calculate \(g(1)\), resulting in \(g(1) = 1^2 = 1\).
• Use 1 as input for \(f(x)\), so \(f(1) = 1 + 3 = 4\).

Thus, \(f \circ g(1) = 4\).
Understanding function composition requires visualizing how function outputs cascade into each other, much like machines on an assembly line.

Function Evaluation

Function evaluation is a fundamental concept where you calculate the value of a function at a specific input. It's about substituting a specific number into a function and simplifying. Let's consider \(f(x) = x + 3\). To evaluate this at \(x = 2\):

• Substitute 2 for \(x\): \(f(2) = 2 + 3\).
• Simplify to find \(f(2) = 5\).

Similarly, to evaluate \(g(x) = x^2\) at \(x = 3\), you substitute 3 and compute \(g(3) = 3^2 = 9\).
This concept helps understand how a function behaves at particular values, forming the basis for more complex operations like composites and algebraic manipulations.

Algebraic Operations on Functions

Algebraic operations on functions include adding, subtracting, multiplying, and dividing functions. These operations are essential for creating new functions based on existing ones. Let's say we have \(f(x) = x + 3\) and \(g(x) = x^2\).
To find \( (f+g)(2) \: \):

• Evaluate each separately: \(f(2) = 5\), \(g(2) = 4\).
• Add them: \(f(2) + g(2) = 5 + 4 = 9\).

As for multiplication like \( (f \cdot g)(0) \: \):

• Compute: \(f(0) = 3\) and \(g(0) = 0\).
• Multiply them to get \(0\).

Performing these operations enhances your ability to manipulate and understand functions significantly.