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Chapter 1: Problem 36

Find (a) \(f \circ g\) and (b) \(g \circ f .\) Find the domain of each function and each composite function. $$f(x)=x^{2}+1, \quad g(x)=\sqrt{x}$$

Short Answer

Expert verified
a) \(f \circ g(x) = x + 1\), domain: \(x \ge 0\). b) \(g \circ f(x) = \sqrt{x^2 + 1}\), domain: all real numbers. Domains of independent functions: \(f(x)\): all real numbers, \(g(x)\): \(x \ge 0\).

Step by step solution

01

Find \(f \circ g\)

Substitute \(g(x)\) into \(f(x)\), yielding: \(f(g(x))= f(\sqrt{x})= (\sqrt{x})^2 + 1 = x + 1\).
02

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

The domain of \(f \circ g\) is the set of all real numbers \(x\) such that \(g(x)\) is defined and \(f(g(x))\) is also defined. Since the function \(g(x)=\sqrt{x}\) is only defined for \(x \ge 0\) and the function \(f(x)= x^2 + 1\) is defined for all real numbers, the domain of \(f \circ g\) is \(x \ge 0\).
03

Find \(g \circ f\)

Substitute \(f(x)\) into \(g(x)\), yielding: \(g(f(x)) = g(x^2+1) = \sqrt{x^2 + 1}\).
04

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

The domain of \(g \circ f\) is the set of all real numbers \(x\) such that \(f(x)\) is defined and \(g(f(x))\) is also defined. Since the function \(f(x)= x^2 + 1\) is defined for all real numbers and the function \(g(x)=\sqrt{x}\) is also defined for all \(x \ge 0\) , the domain of \(g \circ f\) is all real numbers.
05

Find the domains of \(f(x)\) and \(g(x)\)

The domain of \(f(x) = x^2 + 1\) consists of all real numbers, because squaring any real number will produce a real result. The domain of \(g(x) = \sqrt{x}\) consists of all \(x\) such that \(x \ge 0\), because taking the square root of a negative number is undefined in the real number system.

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