Problem 37 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=3-x^{2}, g(x)=x^{2}+2 x-15$$

Step-by-Step Solution

Answer

The function (f+g)(x) equals \(2x-12\), (f-g)(x) equals \(-3x+18\), (fg)(x) equals \(-x^{4}+x^{3}-45x^{2}+30x-45\), and \(\frac{f}{x}\) equals \(\frac{3-x^{2}}{x^{2}+2x-15}\). The domains of the functions (f+g)(x), (f-g)(x), and (fg)(x) are all real numbers. The domain of \(\frac{f}{x}\) is all real numbers except -5 and 3.

Step 1: Add f(x) and g(x)

Add the functions f and g together: \((f+g)(x) = f(x) + g(x) = (3-x^{2}) + (x^{2}+2x-15) = 2x-12\). The domain for this function is all real numbers.

Step 2: Subtract g(x) from f(x)

Subtract g from f: \((f-g)(x) = f(x) - g(x) = (3-x^{2}) - (x^{2}+2x-15) = -3x+18\). The domain for this function is all real numbers.

Step 3: Multiply f(x) and g(x)

Multiply the functions f and g: \((fg)(x) = f(x) \cdot g(x) = (3-x^{2}) \cdot (x^{2}+2x-15) = -x^{4}+x^{3}-45x^{2}+30x-45\). The domain for this function is all real numbers.

Step 4: Divide f(x) by g(x)

\((f/x)(x) = f(x) / g(x) = (3-x^{2}) / (x^{2}+2x-15)\). The domain for this function will exclude values of x that make the denominator zero. So, we solve the equation \(x^{2}+2x-15 = 0\) which gives us x=-5 and x=3. Therefore, the domain is all real numbers except -5 and 3.

Key Concepts

Domain of a Function

When discussing the domain of a function, we're essentially looking at all the possible input values the function can handle without running into any issues.
In the simplest terms, the domain is the set of all values for which a function is defined.

• The domain can sometimes be limited by factors such as division by zero or taking the square root of a negative number in the context of real numbers.
• For polynomial functions, the domain is often all real numbers since polynomials extend infinitely in both directions of the x-axis without causing any calculation errors.
• Rational functions, however, can have more restrictive domains. This is because if a denominator can be reduced down to zero, the function at that point is undefined.

In this exercise, when adding, subtracting, or multiplying the functions, there are no restrictions since you're dealing with straightforward polynomial operations. However, when you divide one function by another, look out for any values that cause the denominator to become zero, thereby restricting the domain at those points.

Polynomial Functions

Polynomial functions form the backbone of many algebraic operations. They consist of terms that are variables raised to whole-number exponents and have coefficients. These functions can be as simple as a line or as complex as a curve with multiple peaks and valleys.

• Expressions like \(3 - x^{2}\) or \(x^{2} + 2x - 15\) are polynomials due to their structure, consisting of variable terms raised to powers.
• Operations such as addition, subtraction, and multiplication on polynomials result in another polynomial because these operations do not change the nature of the expression.
• The degree of a polynomial, which is the highest power of the variable, dictates the overall shape and behavior of its graph.

One of the beauties of polynomial functions is their smooth and continuous nature over their domains, which makes them easier to work with in algebraic operations. They are also defined for all real numbers, so unless something like division is involved, polynomials generally don't face any domain restrictions.

Rational Functions

Rational functions involve the ratio of two polynomials, much like fractions involve numerators and denominators. These functions can sometimes be tricky because the denominator can impose restrictions on the domain.

• A typical rational function might look like \(\frac{f(x)}{g(x)}\), where both \(f(x)\) and \(g(x)\) are polynomials.
• To find problematic points in the domain, you need to find the values that make the denominator zero, as these are where the function becomes undefined.
• In the exercise, \(g(x) = x^{2} + 2x - 15\) was set to zero to find values \(x = -5\) and \(x = 3\) that should be excluded from the domain.

Rational functions can behave quite differently near these points, often shooting off towards infinity or negative infinity. Understanding and identifying these key exclusions ensures that we properly define and handle these functions in a variety of calculations.