Problem 4 - Applied Calculus

Question

Let \(f(x)=x^{2}\) and \(g(x)=3 x-1 .\) Find the following: (a) \(\quad f(2)+g(2)\) (b) \(\quad f(2) \cdot g(2)\) (c) \(\quad f(g(2))\) (d) \(g(f(2))\)

Step-by-Step Solution

Answer

(a) 9; (b) 20; (c) 25; (d) 11.

Step 1: Calculate f(2)

First, we need to evaluate the function \( f(x) = x^2 \) at \( x = 2 \). Substitute 2 into the function: \( f(2) = 2^2 = 4 \).

Step 2: Calculate g(2)

Next, we evaluate the function \( g(x) = 3x - 1 \) at \( x = 2 \). Substitute 2 into the function: \( g(2) = 3(2) - 1 = 6 - 1 = 5 \).

Step 3: find f(2) + g(2)

Now that we have \( f(2) = 4 \) and \( g(2) = 5 \), we add them together: \( f(2) + g(2) = 4 + 5 = 9 \).

Step 4: find f(2) \cdot g(2)

Using the values found, calculate the product \( f(2) \cdot g(2) = 4 \cdot 5 = 20 \).

Step 5: Calculate g(2) for f(g(2))

Recall from Step 2 that \( g(2) = 5 \). We substitute \( 5 \) into the function \( f(x) \).

Step 6: Calculate f(g(2))

Substitute the result of \( g(2) \) into the function \( f(x) = x^2 \): \( f(g(2)) = f(5) = 5^2 = 25 \).

Step 7: Calculate f(2) for g(f(2))

Recall from Step 1 that \( f(2) = 4 \). Substitute \( 4 \) into the function \( g(x) \).

Step 8: Calculate g(f(2))

Substitute the result of \( f(2) \) into the function \( g(x) = 3x - 1 \): \( g(f(2)) = g(4) = 3(4) - 1 = 12 - 1 = 11 \).

Key Concepts

Composite Functions

A composite function is formed when two functions are combined to make a new function. Imagine you have two separate functions: say, a function for making icing and another for baking a cake. When you put them together, you get iced cake, a composite product made from combining two separate processes. In mathematical terms, if you have two functions, \( f(x) \) and \( g(x) \), you can form a composite function by plugging one function into another. This is written as \( (f \circ g)(x) = f(g(x)) \). It's important to note the order matters: \( f(g(x)) \) is not the same necessarily as \( g(f(x)) \).

In the original exercise, we see two composite functions: \( f(g(2)) \) and \( g(f(2)) \). For \( f(g(2)) \), you first find what \( g(2) \) equals and then plug that outcome into \( f(x) \). Similarly, for \( g(f(2)) \), you first calculate \( f(2) \) and substitute that into \( g(x) \). This shows a step-by-step combination of calculations that result in a whole new process reminiscent of our iced cake example.

Algebraic Operations

Algebraic operations are basic computations that include addition, subtraction, multiplication, and division. In functions, these operations can be applied directly to function values once they are evaluated.

Consider the operations \( f(2) + g(2) \) and \( f(2) \cdot g(2) \) from the original exercise. After evaluating \( f(2) \) and \( g(2) \), we simply add the results for part (a) and multiply them for part (b).

• **Addition**: Calculate each function value, then combine with addition, as you would add any normal numbers. This doesn't change the order or process - it's straightforward.
• **Multiplication**: Similarly, find each function value, and then multiply these results together. Consider the rules of multiplication just as you would with integers.

These operations highlight how functions uphold normal numerical rules, following similar patterns and yielding results via basic arithmetic.

Function Notation

Function notation is a shorthand way to express mathematical functions and their evaluations. Think of it like a tool that helps you quickly understand which rule (function) to use and for what input.

In function notation, \( f(x) \) signifies a function named \( f \) with an independent variable \( x \). When you see \( f(2) \), it's asking, "What is the result of the function \( f \) when \( x \) is 2?"

• **Clarity**: Function notation clarifies which operation to use, making it easy to switch and modify different inputs.
• **Convenience**: It simplifies complex calculations by using a standard form, avoiding lengthy expressions.
• **Flexibility**: With function notation, you can conveniently address multiple inputs and functions, adding, subtracting, multiplying, or even composing them.

By using function notation, like \( f(x) = x^2 \) and \( g(x) = 3x - 1 \), these expressions become systematic and easier to manipulate, leading to clear and concise calculation.