Question

Find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((f g)(x)\), and (d) \((f / g)(x)\). What is the domain of \(f / g\) ? \(f(x)=x^{2}+5, \quad g(x)=\sqrt{1-x}\)

Short answer

(a) (f+g)(x) = \(x^{2}+5+ \sqrt{1-x}\), (b) (f-g)(x) = \(x^{2}+5- \sqrt{1-x}\), (c) (fg)(x) = \((x^{2}+5) \cdot \sqrt{1-x}\), (d) (f/g)(x) = \(\frac{x^{2}+5}{\sqrt{1-x}}\) for \(x \in [0,1)\)

Step-by-step solution

Calculate (f+g)(x)

Add the functions \(f(x)\) and \(g(x)\) together: (f+g)(x) = \(f(x) + g(x) = x^{2}+5+ \sqrt{1-x}\)

Calculate (f-g)(x)

Subtract the function \(g(x)\) from \(f(x)\): (f-g)(x) = \(f(x) - g(x) = x^{2}+5- \sqrt{1-x}\)

Calculate (fg)(x)

Multiply \(f(x)\) and \(g(x)\) together: (fg)(x) = \(f(x) \cdot g(x) = (x^{2}+5) \cdot \sqrt{1-x}\)

Calculate (f/g)(x)

Divide \(f(x)\) by \(g(x)\): (f/g)(x) = \(f(x) / g(x) = \frac{x^{2}+5}{\sqrt{1-x}}. Remember, this equation is only valid when \sqrt{1-x} \neq 0\)

Determine the Domain of f/g

Both \(f(x)\) and \(g(x)\) are defined for \(x \in [0,1]\). The denominator \(\sqrt{1-x}\) is zero at \(x=1\), which would make the fraction \((f/g)(x)\) undefined. Therefore, the domain of \((f/g)(x)\) is \(x \in [0,1)\)

Key concepts

Function Addition

When dealing with operations on functions, function addition is one of the foundational concepts. Essentially, adding two functions, let's say function f and function g, involves finding the sum of their values at a particular point x. In our exercise example, you have two functions, f(x) and g(x), defined as f(x) = x^2 + 5 and g(x) = \(\sqrt{1-x}\).

To perform the function addition, you would combine these functions together like this:
\[ (f+g)(x) = f(x) + g(x) = x^{2}+5+ \sqrt{1-x} \].

It simply means for any value of x you plug into f and g, you calculate each function's value and then add those values together to get the result for (f+g)(x).

Function Subtraction

Similar to function addition, function subtraction is another basic operation that can be performed between two functions. In this case, you calculate the difference between the corresponding function values at a specific point x. Referring again to our initial functions, subtracting g(x) from f(x) would give us:
\[ (f-g)(x) = f(x) - g(x) = x^{2}+5 - \sqrt{1-x} \].

This implies that for any given x, you subtract the value of g(x) from f(x) to find the result of (f-g)(x). It's crucial to pay attention to the order of subtraction here, as it affects the outcome (function subtraction is not commutative).

Function Multiplication

Function multiplication is a bit more involved than addition and subtraction but follows a similar principle. When you multiply two functions, you're finding the product of their values at a certain x. So in our example:
\[ (fg)(x) = f(x) \( \cdot \) g(x) = (x^{2}+5) \( \cdot \) \sqrt{1-x} \].

Here, you multiply the value of f(x) with the value of g(x) for the same value of x to get (fg)(x). Remember, the order in which you multiply the functions does not matter (function multiplication is commutative).

Function Division

Function division is often where students need to exercise caution, as it introduces the concept of division by zero, which is undefined. To divide function f by function g at a specific x, you compute the value of f(x) and divide it by the value of g(x), provided that g(x) is not zero at that point.
\[ (f/g)(x) = \frac{f(x)}{g(x)} = \frac{x^{2}+5}{\sqrt{1-x}} \].

As shown in our example, the division is only valid when \(\sqrt{1-x} eq 0\), meaning you cannot include the value of x that makes g(x) zero in the domain of (f/g)(x).

Domain of a Function

The domain of a function is the set of all possible inputs (x-values) for which the function is defined. It is crucial when performing operations such as division because it tells you what values are permissible to plug into the function. In the case of dividing f(x) by g(x), you must exclude any value in the domain that makes the denominator zero, as that would result in an undefined expression.

In our sample problem, the domain of the division function (f/g)(x) excludes the value where g(x) equals zero, in this case, x = 1. Therefore, the domain of (f/g)(x) can be written as:\[ x \in [0,1) \], indicating that x can range from 0 to just below 1.