Problem 56 - Precalculus Essentials

Question

Find. a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=7 x+1, g(x)=2 x^{2}-9$$

Step-by-Step Solution

Answer

a) \((f \circ g)(x) = 14x^{2} - 62\), b) \((g \circ f)(x) = 98x^{2} + 28x - 7\), c) \((f \circ g)(2) = 14\), d) \((g \circ f)(2) = 381\)

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

From the given exercise, identify \(f(x)=7x+1\) and \(g(x)=2x^{2}-9\).

Step 2: Compute \(f \circ g)(x)\)

Substitute \(g(x)\) into \(f(x)\) to find \(f(g(x))\). Therefore, \(f(g(x)) = 7(2x^{2}-9)+1 = 14x^{2}-62.\)

Step 3: Compute \(g \circ f)(x)\)

Now substitute \(f(x)\) into \(g(x)\) to find \(g(f(x))\). Thus, \(g(f(x)) = 2(7x+1)^{2}-9 = 98x^{2}+28x-7.\)

Step 4: Compute \((f \circ g)(2)\)

Now, substituting \(x=2\) into \(f(g(x))=14x^{2}-62\), yields \((f \circ g)(2) = 14(2)^{2}-62 = 14.\)

Step 5: Compute \((g \circ f)(2)\)

Substitute \(x=2\) into \(g(f(x))=98x^{2}+28x-7\) resulting in \((g \circ f)(2)= 98(2)^{2}+28(2)-7 = 381.\)

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers of variables. The variables here are raised to whole number exponents. A polynomial can have constants, variables, and exponents like in the formula: \[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \] where \(a_n, a_{n-1}, \ldots, a_0\) are constants called coefficients and \(n\) is a non-negative integer.In this exercise, we see two polynomial functions:

• \(f(x) = 7x + 1\) which is a linear polynomial because the highest power of \(x\) is 1.
• \(g(x) = 2x^2 - 9\) which is a quadratic polynomial, since the highest power of \(x\) is 2.

Polynomial functions are versatile in mathematics because they can model complex relationships in a simple way while still having precise control over their shape through coefficients and exponents. Understanding these types of functions will provide a foundation for evaluating and manipulating expressions in algebra and calculus.

Evaluation of Functions

Evaluating a function means finding the value of the function's output for a given input value. In mathematical terms, if given a function \(f(x)\), to evaluate it at a specific value \(c\), you simply replace \(x\) with \(c\) in the function's expression. For example, with \(f(x) = 7x + 1\), and you want \(f(2)\), substitute \(2\) into \(f(x)\):\[ f(2) = 7(2) + 1 = 15 \]This process is fundamental because it allows us to calculate and predict values of the function based on input conditions.Similarly, when dealing with the composition of functions, as seen in this exercise, evaluating becomes crucial. Once you find the composite function, say \((f \circ g)(x)\), you'll often need to evaluate it at a specific number, like in step 4. Here, evaluating refers to the substitution and simplification to find the final numerical answer.

Substitution in Functions

Substitution in functions is a technique used to replace a part of an equation or a function with another expression. This process is at the heart of function composition, a crucial concept in algebra.Function composition means creating a new function by inserting one function into another. If you have two functions \(f(x)\) and \(g(x)\), their composition \((f \circ g)(x)\) means plugging \(g(x)\) into \(f(x)\). Mathematically, it looks like:\[ f(g(x)) = f(2x^2 - 9) = 7(2x^2 - 9) + 1 = 14x^2 - 62 \]Here, you substituted the entire expression \(g(x)\) into each instance of \(x\) in \(f(x)\).Substitute operations allow us to handle and simplify complex expressions by dealing with one layer at a time. Practicing substitution enhances problem-solving skills, making handling both simple and intricate equations easier in fields ranging from algebra to calculus and beyond.