Problem 34 - Precalculus : Building Concepts and Connections

Question

In Exercises \(17-40,\) let \(f(x)=-x^{2}+x, g(x)=\frac{2}{x+1},\) and \(h(x)=-2 x+1 .\) Evaluate each of the following. $$(f h)(1)$$

Step-by-Step Solution

Answer

-2

Step 1: Find the function h(1)

Before we can evaluate \((f h)(1)\), we first need to find the value of the function \(h(1)\). Substitute \(x = 1\) into \(h(x) = -2x + 1.\) This would look like \(h(1) = -2(1) + 1 = -2 + 1 = -1.\)

Step 2: Evaluate the function f(h(1))

Next, substitute the result of \(h(1)\), which is \(-1\) into the function \(f(x) = -x^{2} + x\). This would look like: \(f(h(1)) = f(-1) = -(-1)^2 + (-1) = -1 - 1 = -2.\)

Step 3: Simplify the result

After evaluating the composition, we have \(f(h(1))=-2\). There are no further calculations to perform, so \(-2\) is the final answer.

Key Concepts

Polynomial Functions

Polynomial functions are a fundamental concept in mathematics. With these functions, variables are raised to whole number powers. Each polynomial is a sum of terms. These terms have coefficients, which are numbers multiplying the variable raised to a power. For example, in the polynomial function \( f(x) = -x^2 + x \), the term \(-x^2\) has a coefficient of -1 and a power of 2.

Polynomials can take various forms:

• Linear polynomials: These have the form \(ax + b\) and represent straight lines.
• Quadratic polynomials: These take the form \(ax^2 + bx + c\), and their graphs are parabolas.
• Higher-degree polynomials: These involve cubic (like \(x^3\)) or higher powers and have more complex graphs.

Understanding polynomials is crucial for studying more complex mathematical concepts. They appear in algebra, calculus, and various areas of science and engineering.

Function Evaluation

Function evaluation is about finding the output of a function given an input. It is a central concept in mathematics because it helps us see how a function behaves. To evaluate a function, substitute the given input (usually a number) into the function's expression.

For example, to evaluate \(h(1)\) for the function \(h(x) = -2x + 1\), replace the \(x\) with 1. Doing so gives \(-2 \times 1 + 1 = -1\). So, \(h(1)\) is \(-1\). This step shows how specific values behave in broader expressions, key in solving mathematical problems.

Function evaluation also applies to composed functions, like \((f \circ h)(x)\), which we'll explore further. Essentially, you first find the output of the inner function and then use this result as the input for the outer function.

Step-by-Step Solutions

Step-by-step solutions are a methodical way to break down complex problems into understandable pieces. With each step, you refine your understanding and navigate through the problem logically.

Consider the problem of evaluating \((f h)(1)\). Following the steps lays out a clear path:

• First, evaluate the inner function \(h(1)\), giving \(-1\).
• Then, use this result to evaluate the outer function: \(f(-1) = -2\).
• Finally, simplify your calculation where needed, confirming your final answer as \(-2\).

This approach not only helps in arriving at the final solution but builds your mathematical intuition by showing how parts of a problem are interconnected. Each step focuses on one part, making it easier to detect and correct mistakes.