Problem 12 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-14: If \(a\) is a positive real number, find (a) \(g\left(\frac{1}{a}\right)\) (b) \(\frac{1}{g(a)}\) (c) \(g(\sqrt{a})\) (d) \(\sqrt{g(a)}\) $$ g(x)=2 x-5 $$

Step-by-Step Solution

Answer

(a) \(\frac{2}{a} - 5\) (b) \(\frac{1}{2a - 5}\) (c) \(2\sqrt{a} - 5\) (d) \(\sqrt{2a - 5}\)

Step 1: Substitute in g(x) for g(1/a)

To find \(g\left(\frac{1}{a}\right)\), substitute \(\frac{1}{a}\) into the function \(g(x) = 2x - 5\). This gives:\[ g\left(\frac{1}{a}\right) = 2\left(\frac{1}{a}\right) - 5 \]Simplifying, we have:\[ g\left(\frac{1}{a}\right) = \frac{2}{a} - 5 \].

Step 2: Substitute in g(x) for 1/g(a)

To find \(\frac{1}{g(a)}\), first calculate \(g(a)\) using \(g(x) = 2x - 5\). Substitute \(a\) in place of \(x\):\[ g(a) = 2a - 5 \]Then, evaluate \(\frac{1}{g(a)}\):\[ \frac{1}{g(a)} = \frac{1}{2a - 5} \].

Step 3: Substitute in g(x) for g(sqrt(a))

To find \(g(\sqrt{a})\), substitute \(\sqrt{a}\) into the function \(g(x) = 2x - 5\). This gives:\[ g(\sqrt{a}) = 2\sqrt{a} - 5 \].

Step 4: Substitute in g(x) for sqrt(g(a))

To find \(\sqrt{g(a)}\), first evaluate \(g(a)\) as done previously:\[ g(a) = 2a - 5 \]Then, take the square root:\[ \sqrt{g(a)} = \sqrt{2a - 5} \].

Key Concepts

Function evaluation

Function evaluation is a fundamental concept where we substitute a value into a given function to find the output. It's like plugging in different values into a machine to see what comes out. In this exercise, we work with two main operations:

• Substituting \( \frac{1}{a} \) into the function \( g(x) = 2x - 5 \). This requires placing \( \frac{1}{a} \) wherever you see \( x \) in the expression. The calculation becomes \( 2\left( \frac{1}{a} \right) - 5 \), leading to the simplified form \( \frac{2}{a} - 5 \).
• Next, we evaluate \( g(a) \) by substituting \( a \) directly into the function. This process gives \( 2a - 5 \). From there, we can find \( \frac{1}{g(a)} \), which is simply \( \frac{1}{2a-5} \). This last step involves flipping the result of \( g(a) \).

This process forms the basis for understanding any functional forms. It's a repeatable practice that adds clarity and measurability to otherwise abstract algebraic concepts.
Understanding different operations, such as substitution and evaluation, helps to solve complex algebraic problems efficiently.

Algebraic manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them more understandable or to solve them.In this exercise, we manipulate the function \( g(x) = 2x - 5 \) in various scenarios:

• For instance, simplifying \( g\left( \frac{1}{a} \right) \) to \( \frac{2}{a} - 5 \) involves expanding and combining terms.
• Similarly, calculating \( \frac{1}{g(a)} = \frac{1}{2a-5} \) involves understanding how inverse functions work. Here, the key is recognizing that \( 1 \) divided by any function requires you to invert the result after substituting values.

These manipulations develop a more robust understanding of transforming and reformulating algebraic expressions.
Such skills are crucial when solving real-life problems and in advanced mathematics, where complex expressions need simplification for solutions or interpretations.

Square root operations

Square root operations often appear in algebra, involving roots of numbers or expressions. It's essential to learn how to handle these operations comfortably.In our exercise, we encounter square root operations in two tasks:

• First, \( g(\sqrt{a}) \, = \, 2\sqrt{a} - 5 \), which shows evaluating a function where the input itself is a square root. Here, the nested root doesn't change the process much; we still substitute as usual and simplify the result.
• Secondly, \( \sqrt{g(a)} \, = \, \sqrt{2a - 5} \) emphasizes taking the root of the entire function output, which requires checking if that expression under the root is valid (non-negative) in the real number system.

Square root operations are delicate as they often require ensuring the expressions under the root are positive or zero, ensuring results remain within the real number system.
Mastering these operations enhances fluency in algebra, especially when tackling quadratic equations and functions.