Problem 27 - 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. $$(h-f)(-1)$$

Step-by-Step Solution

Answer

The value of \((h-f)(-1)\) is 5.

Step 1: Find \(h(-1)\)

We need to replace \(x\) in \(h(x)=-2x+1\) with \(-1\). Putting \(-1\) in place of \(x\) in \(h(x)\), we get \(h(-1) = -2*-1+1=2+1=3\)

Step 2: Find \(f(-1)\)

Next, replace \(x\) in \(f(x)=-x^{2}+x\) with \(-1\). Substituting \(-1\) in place of \(x\) in \(f(x)\), we get \(f(-1) = -(-1)^{2} + -1 = -1+ -1= -2\)

Step 3: Evaluate \((h-f)(-1)\)

The last step will involve evaluating \((h-f)(-1)\), which means subtracting the value obtained for \(f(-1)\) from that of \(h(-1)\). So, \((h-f)(-1)= h(-1) - f(-1) = 3 - (-2) = 5

Key Concepts

Polynomial Functions

Polynomial functions are one of the most fundamental concepts in algebra. They are expressions that involve variables raised to whole number powers. Each polynomial is characterized by its terms, where each term is a product of a constant coefficient and a power of the variable. For example, in the function \(f(x) = -x^2 + x\),

• \(-x^2\) is a term with a power of 2, making it a quadratic term.
• \(x\) is a term with a power of 1, which is a linear term.

These kinds of functions can be classified by their degree, which is the highest power of the variable present in the polynomial. Quadratic functions like \(-x^2 + x\) have a degree of 2, meaning they have a parabolic graph.
Polynomial functions can take many shapes depending on their degree and coefficients. They are smooth and continuous, making them very useful in modeling a wide range of real-world situations.

Linear Functions

Linear functions are simpler than polynomial functions due to their first-degree nature. They can be expressed in the form \(y = mx + c\). Here, \(m\) represents the slope, and \(c\) is the y-intercept.

• In the function \(h(x) = -2x + 1\), the slope \(-2\) tells us that for every unit increase in \(x\), the value of \(h(x)\) decreases by 2.
• The y-intercept, \(1\), is the point where the line crosses the y-axis.

Linear functions graph as straight lines and are foundational in understanding more complex functions. They're used to model relationships with constant rate changes, such as speed or cost over time. Simplifying complex scenarios into linear models can make it easier to derive meaningful insights.

Domain and Range

Understanding the domain and range is crucial for function evaluation. The domain of a function is essentially all possible input values (\(x\)-values) that can be plugged into the function. The domain defines where the function is applicable and valid.

• For \(g(x) = \frac{2}{x+1}\), the domain excludes \(x = -1\) since it would result in division by zero.
• Other functions like \(f(x) = -x^2 + x\) generally have a domain of all real numbers because there's no restriction on squaring or adding real numbers.

The range of a function, on the other hand, is the set of possible output values (\(y\)-values) that the function can produce.

• For instance, the range of a quadratic like \(-x^2 + x\) is all real numbers \(y\) that the parabola can reach.
• The range of a linear function like \(-2x + 1\) is all real numbers due to its continuous linear increase or decrease.

Knowing the domain and range provides a comprehensive grasp of where and how functions behave and interact.