Problem 78 - Elementary and Intermediate Algebra

Question

Let \(h(t)=\sqrt{t+3}\) and \(k(t)=t-5 .\) Find each of the following. $$ (h \circ k)(11) $$

Step-by-Step Solution

Answer

\((h \circ k)(11) = 3\).

Step 1: Understand the Problem

We need to find \((h \circ k)(11)\), which means we have to find the function composition \(h(k(11))\). This involves substituting 11 into the function \(k(t)\) and then using that result as the input for the function \(h(t)\).

Step 2: Substitute into the Inner Function

Begin by finding \(k(11)\). Substitute \(t = 11\) into \(k(t) = t - 5\): \[k(11) = 11 - 5 = 6\] So, \(k(11) = 6\).

Step 3: Substitute into the Outer Function

Next, use the result from the previous step as the input for \(h(t)\). Substitute \(t = 6\) into \(h(t) = \sqrt{t+3}\): \[h(6) = \sqrt{6 + 3} = \sqrt{9} = 3\] So, \(h(6) = 3\).

Step 4: Write the Final Answer

Therefore, the composition \((h \circ k)(11) = h(k(11)) = h(6) = 3\).

Key Concepts

functions

In mathematics, functions are like special machines that transform an input into an output. They follow predefined rules for this transformation. When we talk about functions, we refer to relationships where each input (often called "t") is linked to exactly one output. For example, in our exercise, we have two functions:

• Function \( h(t) = \sqrt{t+3} \) applies the square root to \( t+3 \).
• Function \( k(t) = t - 5 \) simply subtracts 5 from \( t \).

Both functions begin with an input, perform computations, and give us an output. When dealing with functions, we often write them as \( f(t) \), \( g(t) \), or in this case, \( h(t) \) and \( k(t) \). Understanding how to use these "machines" to determine the correct output is essential in mathematics, particularly in algebra.

algebraic manipulation

Algebraic manipulation is a powerful tool that simplifies and solves mathematical expressions and equations. In the context of function composition, this manipulation involves substituting values into functions in a specific sequence. Here's how it works:- We start with function \( k(t) = t-5 \) and substitute \( t = 11 \) into it. The calculation becomes: \[ k(11) = 11 - 5 = 6 \] - We then take the result \( 6 \) and plug it into the next function \( h(t) = \sqrt{t+3} \). This results in: \[ h(6) = \sqrt{6 + 3} = \sqrt{9} = 3 \]Each step requires precise substitution and computation to ensure accuracy. By mastering algebraic manipulation, students can break down complex problems into simpler parts, making them more manageable and less daunting.

square root function

The square root function is a type of mathematical function that "reverses" the process of squaring a number. The square root of a number gives the original number that was squared. Most of the time, we write the square root function as \( \sqrt{ } \) . In our exercise, the function \( h(t) = \sqrt{t+3} \) directly uses this operation. When working with square root functions, here are a few important points to remember:

• The operation \( \sqrt{ } \) is only defined for non-negative numbers in the real number system because square roots of negative numbers aren't real (here, we work within real numbers).
• Square roots reverse the squaring process. If we have \( \sqrt{9} \), it returns 3, since \( 3^2 = 9 \).
• When substituting into a square root function, perform all arithmetic inside (like addition or subtraction) before applying the square root.
• Simplifying \( \sqrt{x} \) involves recognizing perfect squares (like 9, 16, 25, etc.) when they appear.

Combining these properties allows us to correctly evaluate and simplify expressions involving square roots, such as finding \( h(6) = 3 \) in this exercise.