Problem 80 - Intermediate Algebra

Question

Let \(h(t)=\sqrt{t+3}\) and \(k(t)=t-5 .\) Find each of the following. $$ (k \circ h)(-2) $$

Step-by-Step Solution

Answer

\((k \circ h)(-2) = -4\).

Step 1: Understand the Function Composition

The notation \((k \circ h)(x)\) represents the composition of the functions \(k(x)\) and \(h(x)\). This means you first apply the function \(h(x)\) and then apply \(k(x)\) to the result. Mathematically, it is expressed as \(k(h(x))\).

Step 2: Calculate \(h(-2)\)

Substitute \(-2\) into the function \(h(x) = \sqrt{x + 3}\). \[ h(-2) = \sqrt{-2 + 3} = \sqrt{1} = 1 \]

Step 3: Calculate \(k(h(-2))\)

Now substitute the result from Step 2 into the function \(k(x) = x - 5\). Since we found \(h(-2) = 1\), substitute this into \(k(x)\):\[ k(1) = 1 - 5 = -4 \]

Step 4: Write the Final Answer

The result of the function composition \((k \circ h)(-2)\) is \(-4\).

Key Concepts

Intermediate Algebra

Intermediate Algebra is a branch of mathematics that deals with solving equations and understanding various types of functions. It is a step above basic algebra and introduces more complex concepts but still remains foundational. When studying Intermediate Algebra, you will encounter different function operations, such as function composition.

• Function Composition: This is a key concept in Intermediate Algebra where you combine two functions. You apply one function to an argument and then apply a second function to the result of the first. It's often denoted with a small circle, like \((f \circ g)(x)\), meaning \(f(g(x))\).
• Working with Square Roots: Solving equations often involves applying operations like squaring or taking square roots, which are integral parts of Intermediate Algebra.
• Application of Concepts: Using operations, such as composition, helps in breaking down complicated problems into simpler parts.

Mastering Intermediate Algebra involves familiarizing yourself with a variety of functions and operations, which ultimately prepares you for more advanced mathematical challenges.

Square Root Function

The square root function is fundamental in algebra and particularly significant in Intermediate Algebra. This function can be expressed as \(f(x) = \sqrt{x}\). When you deal with square root functions, you perform the operation of finding a number that, when multiplied by itself, gives the original number.

• Understanding the Domain: The domain of the square root function \(\sqrt{x}\) is all non-negative numbers \(x \geq 0\). This is because there are no real square roots for negative numbers within the set of real numbers.
• Function Operations: In the context of function composition, a square root function can be the inner function, meaning it gets executed first. For instance, in the problem, \(h(t) = \sqrt{t+3}\) was applied first before applying the linear function \(k(t)\).

Square root functions come into play in various calculations and are crucial when performing operations like solving quadratic equations or simplifying expressions.

Linear Function

Linear functions are one of the simplest, yet one of the most important types of functions studied in algebra. They describe a constant rate of change, forming a straight line when graphed. In general, a linear function is expressed as \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

• Slope and Intercept: The slope \(m\) tells you how steep the line is, while the y-intercept \(b\) indicates where the line crosses the y-axis.
• Linear Function in Composition: Linear functions are often used in function composition, as seen with \(k(t) = t - 5\). Despite their simplicity, they play a crucial role in simplifying problems when combined with other types of functions.
• Applications: In real life, linear functions can model relationships where there is a constant change rate, such as speed = distance/time or total cost = price per item times number of items.

These functions are easy to work with, making them a staple in all algebra courses, and an essential tool for modeling and solving a wide range of problems.