Problem 38 - Precalculus with Limits

Question

In Exercises 37-52, evaluate the function at each specified value of the independent variable and simplify. \(g(y) = 7 - 3y\) (a) \(g(0)\) (b) \(g(\frac{7}{3})\) (c) \(g(s+2)\)

Step-by-Step Solution

Answer

The function \(g(y) = 7 - 3y\) evaluates to \(7\) at \(y=0\), to \(0\) at \(y=\frac{7}{3}\) and to \(1-3s\) at \(y=s+2\).

Step 1: Evaluate at \(y=0\)

Substitute \(y=0\) into the expression for \(g(y)\): \(g(0) = 7 - 3(0) = 7 - 0 = 7\).

Step 2: Evaluate at \(y=\frac{7}{3}\)

Substitute \(y=\frac{7}{3}\) into the expression for \(g(y)\): \(g(\frac{7}{3}) = 7 - 3(\frac{7}{3}) = 7 - 7 = 0\).

Step 3: Evaluate at \(y=s+2\)

Substitute \(y=s+2\) into the expression for \(g(y)\): \(g(s+2) = 7 - 3(s+2) = 7 - 3s - 6 = 1 - 3s\).

Key Concepts

Function Evaluation

When studying mathematics, the concept of function evaluation is a fundamental skill that you will often utilize. Evaluating a function involves finding the value of the function's output, known as the dependent variable, for a particular value of the input, or the independent variable. This process is essentially substituting the given input value into the function and simplifying the result.

For example, given the function from the exercise, \(g(y) = 7 - 3y\), evaluating \(g(0)\) means finding the value of \(g\) when \(y\) is 0. By substituting 0 for \(y\) in the function, we get \(g(0) = 7 - 3(0) = 7\), which simplifies to 7. Understanding this process is crucial as it forms the basis of more advanced topics in algebra and calculus.

Linear Functions

In algebra, linear functions are one of the simplest forms of functions you can encounter. These functions have the characteristic of producing a straight line when graphed on a coordinate plane. They follow the format \(f(x) = mx + b\), where \(m\) represents the slope, which indicates the steepness of the line, and \(b\) represents the y-intercept, the point where the line crosses the y-axis.

The function given in our exercise, \(g(y) = 7 - 3y\), is an example of a linear function. Here, the slope is -3, and the y-intercept is 7. This illustrates that for every unit increase in the independent variable \(y\), the value of \(g(y)\) will decrease by 3 units, a vital concept when interpreting and graphing linear relationships.

Algebraic Substitution

The technique of algebraic substitution is a method widely used in evaluating functions. It involves replacing a variable in an expression with a given number or another expression. This method is particularly useful when dealing with functions and formulas. It reduces complex problems into simpler ones which can then be easily simplified or solved.

Taking a closer look at part (c) from the exercise, we apply algebraic substitution to evaluate \(g(s+2)\). Here, we replace the \(y\) in the function with \(s+2\), resulting in \(g(s+2) = 7 - 3(s+2)\). Expanding this gives us \(7 - 3s - 6\), which then simplifies to \(1 - 3s\). Mastering algebraic substitution is not only useful in evaluating functions but is also essential in solving equations, performing integrations, and simplifying algebraic expressions.