Question

Find the domains of \(f \circ g\) and \(g \circ f\). (a) \(f(x)=\sqrt{x}, \quad g(x)=1-x^{2}\) (b) \(f(x)=\log x, \quad g(x)=\sin x\) (c) \(f(x)=\frac{1}{1-x^{2}}, \quad g(x)=\cos x\) (d) \(f(x)=\sqrt{x}, \quad g(x)=\sin 2 x\)

Short answer

(a) \(f(x)=\sqrt{x}, \quad g(x)=1-x^{2}\) \(f \circ g\) domain: \(-1 \leq x \leq 1\) \(g \circ f\) domain: \(x \geq 0\) (b) \(f(x)=\log x, \quad g(x)=\sin x\) \(f \circ g\) domain: \((2n\pi, (2n+1)\pi)\) where n is an integer \(g \circ f\) domain: \(x > 0\) (c) \(f(x)=\frac{1}{1-x^{2}}, \quad g(x)=\cos x\) \(f \circ g\) domain: all real numbers, except \(x=2n\pi\) and \(x=(2n+1)\pi\) where n is an integer \(g \circ f\) domain: all real numbers except \(x=\pm1\) (d) \(f(x)=\sqrt{x}, \quad g(x)=\sin 2 x\) \(f \circ g\) domain: \([n\pi-\frac{\pi}{4}, n\pi+\frac{\pi}{4}]\) where n is an integer \(g \circ f\) domain: \(x \geq 0\)

Step-by-step solution

Identify the domains of f and g

The domain of \(f(x)=\sqrt{x}\) is \(x \geq 0\) since the square root of a negative number is undefined. The domain of \(g(x)=1-x^2\) is all real numbers.

Find the domain of \(f \circ g\)

For the composition \(f(g(x))=\sqrt{1-x^2}\), we must find the values of x for which \(1-x^2 \geq 0\). Solve the inequality: \(1-x^2 \geq 0 \implies x^2 \leq 1 \implies -1 \leq x \leq 1\) Thus, the domain for \(f \circ g\) is \(-1 \leq x \leq 1\).

Find the domain of \(g\circ f\)

For the composition \(g(f(x))=1-(\sqrt{x})^2 = 1-x\), the domain is all the values for which x is non-negative since f(x) can only take non-negative values. Thus, the domain for \(g \circ f\) is \(x \geq 0\). (b) \(f(x)=\log x, \quad g(x)=\sin x\)

Identify the domains of f and g

The domain of \(f(x)=\log x\) is \(x > 0\) since logarithms are defined only for positive numbers. The domain of \(g(x)=\sin x\) is all real numbers.

Find the domain of \(f \circ g\)

Evaluate \(f(g(x))=\log(\sin x)\). The log function requires a positive input value. Since the sine function value ranges from -1 to 1, we need \(0 < \sin x < 1\). This occurs for intervals of the form \((2n\pi,\ (2n+1)\pi)\), where n is an integer. Thus, the domain for \(f \circ g\) is \((2n\pi,\ (2n+1)\pi)\).

Find the domain of \(g\circ f\)

Evaluate \(g(f(x))=\sin(\log x)\). Since the sine function accepts all real numbers, the only constraint on the domain of \(g\circ f\) is the domain of \(f(x)\). Thus, the domain for \(g \circ f\) is \(x > 0\). (c) \(f(x)=\frac{1}{1-x^{2}}, \quad g(x)=\cos x\)

Identify the domains of f and g

The domain of \(f(x)=\frac{1}{1-x^2}\) is all the values except \(x=\pm1\), since division by zero is not allowed. The domain of \(g(x)=\cos x\) is all real numbers.

Find the domain of \(f \circ g\)

Evaluate \(f(g(x))=\frac{1}{1-\cos^2 x}\). Since cosine ranges from -1 to 1, the only values of x that would make the denominator zero are when \(\cos x = \pm1\). This occurs at \(x=2n\pi\) and \(x=(2n+1)\pi\) where n is an integer. Thus, the domain for \(f \circ g\) is all real numbers, except the above mentioned values.

Find the domain of \(g\circ f\)

Evaluate \(g(f(x))=\cos\left(\frac{1}{1-x^2}\right)\). Since the cosine function accepts all real numbers, the only constraint on the domain of \(g\circ f\) is the domain of \(f(x)\). Thus, the domain for \(g \circ f\) is all real numbers except \(x=\pm1\). (d) \(f(x)=\sqrt{x}, \quad g(x)=\sin 2 x\)

Identify the domains of f and g

The domain of \(f(x)=\sqrt{x}\) is \(x \geq 0\) since the square root of a negative number is undefined. The domain of \(g(x)=\sin 2x\) is all real numbers.

Find the domain of \(f \circ g\)

Evaluate \(f(g(x))=\sqrt{\sin 2x}\). The square root requires a non-negative input value. Since the sine function value ranges from -1 to 1, we need \(0 \leq \sin 2x \leq 1\). This occurs for intervals of the form \([n\pi-\frac{\pi}{4},\ n\pi+\frac{\pi}{4}]\), where n is an integer. Thus, the domain for \(f \circ g\) is \([n\pi-\frac{\pi}{4},\ n\pi+\frac{\pi}{4}]\).

Find the domain of \(g\circ f\)

Evaluate \(g(f(x))=\sin(2\sqrt{x})\). The sine function accepts all real numbers, and the only constraint on the domain of \(g\circ f\) is the domain of \(f(x)\). Thus, the domain for \(g \circ f\) is \(x \geq 0\).

Key concepts

Domain of a Function

The domain of a function is a crucial concept in mathematics as it defines the set of all input values (usually denoted as \(x\)) for which the function produces a valid output. To determine the domain, one must consider any restrictions imposed by operations within the function, such as division by zero, logarithms, and square roots. For example:

• In the function \(f(x) = \frac{1}{x}\), the domain is all real numbers except \(x = 0\) because division by zero is undefined.
• For \(f(x) = \log(x)\), the domain is \(x > 0\) since logarithms require positive inputs.

Understanding the domain is essential when dealing with function composition, as the output of one function becomes the input of another, thereby combining the domain restrictions of both functions.

Square Root Function

The square root function is represented as \(f(x) = \sqrt{x}\). It is a fundamental function in algebra and has a specific domain requirement: the input value \(x\) must be non-negative (i.e., \(x \geq 0\)). This is because taking the square root of a negative number results in an undefined (or imaginary) value in real number mathematics.

• When considering the square root function in composition, such as \(f(g(x)) = \sqrt{g(x)}\), we must ensure that \(g(x) \geq 0\).
• This might restrict the permissible values for \(x\) in the domain of the composed function, depending on \(g(x)\).

Identifying these conditions is essential for determining the valid input range for composed functions involving square roots.

Logarithmic Function

Logarithmic functions, such as \(f(x) = \log x\), arise frequently in various branches of mathematics, including calculus and complex analysis. Their domain is strictly \(x > 0\) because the logarithm of zero or a negative number is not defined in the real number system.

• This characteristic plays an important role in function compositions like \(f(g(x)) = \log(g(x))\), where we must ensure \(g(x) > 0\).
• For example, if \(g(x) = \sin x\), then the domain of \(f(g(x))\) involves values of \(x\) such that \(0 < \sin x \leq 1\).

Logarithmic functions are vital in solving exponential equations and understanding these domain restrictions helps avoid undefined or complex outputs.

Trigonometric Function

Trigonometric functions like \(f(x) = \sin x\) or \(g(x) = \cos x\) oscillate between specific ranges. The sine and cosine functions, for instance, range between -1 and 1 and accept all real numbers in their domain.

• For compositions such as \(f(g(x)) = \sin(g(x))\), it is crucial to check the domain of \(g(x)\) first, as it dictates input values for \(f(x)\).
• Some compositions require additional considerations, like avoiding specific values of \(x\) that lead to undefined conditions. For example, \(\cos x\) should never be ±1 if used as a denominator in functions.

Trigonometric identities and periodicity can further affect domain matters, making it necessary to consider their properties when working with compositions.