Problem 66 - Precalculus : Building Concepts and Connections

Question

In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(h \circ f)\left(\frac{3}{2}\right)$$

Step-by-Step Solution

Answer

-11.25

Step 1: Understanding Function Composition

In function composition, when you see something like (h ∘ f)(x), this means you should first apply function f to x, then apply function h to the result. So, \( (h∘f)(x) = h(f(x))\).

Step 2: Substitute Function f into Function h

First, we apply function f to x, \(f(x) = x^{2} + x\). Then, we substitute f(x) into h, giving us \(h(f(x)) = -3(f(x)) = -3(x^{2} + x)\).

Step 3: Substitute Given Value in Function

Next, we substitute \(x = 3/2\) into our function \(h(f(x)) = -3((3/2)^{2} + (3/2)).\)

Step 4: Simplify The Function

Simplify the function to get the numerical value. Perform the operations to solve the function, which will yield \(-3((3/2)^{2} + (3/2)) = -3(9/4 + 3/2) = -3(2.25 + 1.5) = -3(3.75) = -11.25\)

Key Concepts

Understanding Composite Functions

In math, composite functions are like building with blocks. You take one function and plug it into another. Think of it as following a recipe where you start with a basic ingredient, apply the first set of steps, and then use the result in the next part of your recipe. For example, if we have two functions, say \(f(x)\) and \(h(x)\), we can create a composite function by applying \(f\) first, and then \(h\) to the result. This is written as \((h \circ f)(x)\). Simply put, it means "first \(f\), then \(h\)."

Here's how it works:

• You begin with \(x\), a number or variable that you apply to the first function \(f\).
• Next, you plug the result you get from \(f(x)\) into the second function \(h\).
• Finally, conduct any further computations or operations required by \(h(x)\).

This concept is especially useful in solving more complex mathematical problems, as it allows you to break down those problems into simpler, manageable smaller parts. Imagine it as a step-by-step journey, where each function leads you a little closer to your final destination.

Exploring Mathematical Operations in Composite Functions

Mathematical operations within composite functions involve a series of steps to combine and transform initial values. When evaluating composite functions, such as \((h \circ f)(x)\), several operations come into play:

• Substitution: Determine what \(f(x)\) is by plugging \(x\) into the function \(f\). In our exercise, \(f(x) = x^2 + x\).
• Multiplication: Once \(f(x)\) is substituted into \(h(x)\), another operation takes place. For example, substituting \(f(x)\) into \(h(x) = -3x\) results in \(h(f(x)) = -3(x^2 + x)\).
• Calculation: Now, follow the order of operations (often known by the acronym PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction). For example, when plugging \(x = \frac{3}{2}\) into \(h(f(x)) = -3((\frac{3}{2})^2 + \frac{3}{2})\), a series of multiplications and additions take place to simplify the expression.

These mathematical operations are crucial for finding the correct result. By understanding and applying each correctly, students can unravel the complexity behind composite functions and solve them accurately.

Tackling Precalculus Problems

Precalculus serves as a bridge to more advanced mathematical studies and demands a solid understanding of various concepts. Tackling precalculus problems involving composite functions is a stepping stone to mastering calculus.

Understanding function composition is essential, as it lays the groundwork for dealing with more complex functions. In precalculus, problems often involve multiple functions and demand precision in manipulation.

For example, consider our exercise with \(f(x) = x^2 + x\) and \(h(x) = -3x\). To solve a precalculus composite function problem, students need to:

• Ensure clarity on each function and its role.
• Perform each operation carefully to avoid errors, such as skipping steps or mistaking orders of operations.
• Focus on simplifying expressions as much as possible to make calculations easier and reduce the chance for mistakes.

Precalculus problems often teach students to appreciate the logical flow and importance of each step in problem-solving. Mastery in these areas helps students gain confidence in their ability to progress into calculus and beyond.