Problem 3 - Precalculus Mathematics for Calculus

Question

(a) Which of the following functions have 5 in their domain? $$f(x)=x^{2}-3 x \quad g(x)=\frac{x-5}{x} \quad h(x)=\sqrt{x-10}$$ (b) For the functions from part (a) that do have 5 in their domain, find the value of the function at 5.

Step-by-Step Solution

Answer

Functions f(x) and g(x) include 5 in their domains. At x=5, f(x)=10 and g(x)=0.

Step 1: Analyze Function f(x)

The function is given by \( f(x) = x^2 - 3x \). This is a quadratic function and is defined for all real numbers. Therefore, 5 is in the domain of \( f(x) \).

Step 2: Analyze Function g(x)

The function is \( g(x) = \frac{x-5}{x} \). The denominator forbids zero, as division by zero is undefined. Thus, 5 is in the domain of \( g(x) \) because the denominator does not become zero at \( x = 5 \).

Step 3: Analyze Function h(x)

The function is \( h(x) = \sqrt{x-10} \). The expression under the square root must be non-negative, which means \( x - 10 \geq 0 \) or \( x \geq 10 \). Therefore, 5 is not in the domain of \( h(x) \).

Step 4: Evaluate f(x) at x = 5

Since 5 is in the domain of \( f(x) \), we compute \( f(5) = 5^2 - 3 \times 5 = 25 - 15 = 10 \).

Step 5: Evaluate g(x) at x = 5

Since 5 is in the domain of \( g(x) \), we compute \( g(5) = \frac{5 - 5}{5} = \frac{0}{5} = 0 \).

Key Concepts

Quadratic function

A quadratic function is one of the most common types of polynomial functions, characterized by its highest exponent of the variable being two. The general form of a quadratic function is given by \( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants with \( a eq 0 \). The domain of any quadratic function is all real numbers because polynomials are defined for every real number.

So, any value you substitute into the variable \( x \) will yield a result. This is why, for the function \( f(x) = x^2 - 3x \), 5 is included in its domain. The function will yield a calculation without any restriction at \( x = 5 \).

To evaluate the function at \( x = 5 \), plug 5 into the function:

• Compute \( f(5) = 5^2 - 3(5) = 25 - 15 = 10 \)

Thus, the function \( f(x) \) evaluates to 10 at \( x=5 \). This showcases how quadratic functions operate smoothly for all inputs, including 5.

Rational function

Rational functions are functions of the form \( g(x) = \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials. The domain includes all real numbers except those values which make the denominator zero, as division by zero is undefined.

For example, consider the function \( g(x) = \frac{x-5}{x} \). Here, it is important to evaluate the denominator. The critical point is when \( x = 0 \), which would cause the denominator \( x \) to become zero. Hence, \( x eq 0 \) belongs to the domain.

As 5 does not make the denominator zero, it is within the domain.

• To find the value of \( g(x) \) at \( x = 5 \), calculate: \( g(5) = \frac{5 - 5}{5} = \frac{0}{5} = 0 \).

This indicates that rational functions can be defined at most numbers, except where their denominators zero out.

Square root function

A square root function involves taking the square root of an expression, typically written as \( h(x) = \sqrt{x} \). The domain of these functions is limited to those numbers that do not result in a negative expression inside the square root because square roots of negative numbers are not considered real.

For the function \( h(x) = \sqrt{x-10} \), the expression inside the square root must be non-negative:

• Set up the inequality: \( x - 10 \geq 0 \)
• Solve for \( x \): \( x \geq 10 \)

This shows that only values greater than or equal to 10 are in the domain. Since 5 is less than 10, it is excluded from the domain of \( h(x) \). Thus, the expression \( \sqrt{x-10} \) is not defined for \( x = 5 \). This is how square root functions restrict their input values based on maintaining non-negative radicands.