Problem 25 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

If \(h(x)=f(g(x))\), then \(x\) is in the domain of \(h\) if and only if \(x\) is in the domain of \(g\) and \(g(x)\) is in the domain of \(f .\) In other words, \(x\) must be a valid input for \(g\) and \(g(x)\) must be a valid input for \(f\). (a) If \(h(x)=f(g(x))\), where \(g(x)=\sqrt{x}\) and \(f(x)=x^{2}\), what is the largest possible domain of \(h\) ? For all \(x\) in its domain, \(h(x)=x\). Why is the domain not \((-\infty, \infty)\) ? (b) If \(h(x)=f(g(x))\), where \(g(x)=\frac{1}{x-1}\) and \(f(x)=\frac{1}{x+3}\), what is the largest possible domain of \(h ?\) (There are two numbers that must be excluded from the domain.)

Step-by-Step Solution

Answer

(a) The largest possible domain of \(h\) for the first set of functions is \([0, \infty)\). (b) The largest possible domain of \(h\) for the second set of functions is \((-\infty, -3) \cup (-3, 1) \cup (1, \infty)\).

Step 1: Evaluating Function (a)

In the part (a), \(g(x)=\sqrt{x}\) and \(f(x)=x^{2}\). The domain of \(g(x)\) is \([0, \infty)\) because you can't take the square root of a negative number. The output of \(g(x)\) is also \([0, \infty)\), which is the input set for \(f(x)\). The function \(f(x)\) has a domain of \((-\infty, \infty)\). However, since its input only consists of the positive real numbers and zero from \(g(x)\), its effective domain here is also \([0, \infty)\). Thus, the largest possible domain of \(h(x)\) in this case is \([0, \infty)\). The domain is not \((-\infty, \infty)\) because \(x\) must be a valid input for \(g\), and negative values are not permissible.

Step 2: Evaluating Function (b)

In part (b), \(g(x)=\frac{1}{x-1}\) and \(f(x)=\frac{1}{x+3}\). Let's first consider \(g(x)\). The denominator of \(g(x)\) cannot be zero, so \(x≠1\). Thus, the domain for \(g(x)\) is \((-\infty, 1) \cup (1, \infty)\). Thereafter, the output from \(g(x)\), when \(x≠1\), becomes the input for \(f(x)\), which has a domain of \((-\infty, -3) \cup (-3, \infty)\). As a result, the values that need to be excluded from the input of \(f(x)\) are \(x=1\) (from \(g(x)\)) and \(x=-3\) (from \(f(x)\)), meaning the domain of \(h(x)\) is \((-\infty, -3) \cup (-3, 1) \cup (1, \infty)\).

Step 3: Conclusion

By analysing the domain restrictions of \(f(x)\) and \(g(x)\) in both cases, we can deduce the respective domains of \(h(x)\).

Key Concepts

Understanding Function Domain

When we talk about the domain of a function, we mean the set of all possible input values (usually "x") that the function can accept without running into any mathematical trouble.
For example, in the function \(g(x) = \sqrt{x}\), the domain is \([0, \infty)\), because we can only take the square root of non-negative numbers. Negative values would lead to imaginary numbers, which we don't consider here.

For composite functions, like \(h(x) = f(g(x))\), we need to ensure that the output from \(g(x)\) is a valid input for \(f(x)\). If \(f(x)\) is defined everywhere, its domain is \((-\infty, \infty)\), but practically, it only accepts what \(g(x)\) outputs. Therefore, correctly understanding the domain of each function individually and together is crucial.

Exploring the Range of Functions

The range of a function is all the possible output values the function can produce. Consider \(f(x) = x^2\): the range is \([0, \infty)\) because squaring any real number results in a non-negative number.

In composite functions, understanding the range is essential because it helps determine what inputs are valid overall. For instance, with \(h(x) = f(g(x))\), the range of \(g(x)\) must fit into the domain of \(f(x)\). This ensures that the outputs of \(g(x)\) can be legally processed by \(f(x)\), producing final outputs within \(h(x)'s\) range.

When \(g(x) = \sqrt{x}\) and \(f(x) = x^2\), the range of \(g(x)\) is \([0, \infty)\), which is perfect as it's already suitable for \(f(x)\). That's why \(h(x) = x\) remains with a range \([0, \infty)\).

Handling Domain Restrictions

Domain restrictions are specific values that are excluded from a function's domain to avoid undefined situations, like dividing by zero. In \(g(x) = \frac{1}{x-1}\), the function becomes undefined when \(x = 1\), because you can't divide by zero. Therefore, \(x = 1\) is a domain restriction.

For composite functions, multiple domain restrictions can emerge. Take \(h(x) = f(g(x))\) where \(g(x) = \frac{1}{x-1}\) and \(f(x) = \frac{1}{x+3}\). The domain of \(g(x)\) excludes \(x = 1\), and the range of \(g(x)\), as input to \(f(x)\), excludes \(x = -3\).

• In \(g(x)\), avoid \(x = 1\).
• In \(f(g(x))\), avoid \(x = -3\) for inputs from \(g(x)\).

This means the largest domain for \(h(x)\) is \((-\infty, -3) \cup (-3, 1) \cup (1, \infty)\). Recognizing and managing these restrictions ensures functions remain well-defined and computations valid.