Mathematics

Mathematics

Problem

The functions ff and gg are defined as follows: f(x)={1x−1,x<2,6−x,x≥2,g(x)={log⁡e(x−3),x>3,x2,x≤3.f(x)=\begin{cases} \dfrac{1}{x-1}, & x<2,\\ \sqrt{6-x}, & x\ge 2, \end{cases} \quad g(x)=\begin{cases} \log_{e}(x-3), & x>3,\\ x^{2}, & x\le 3. \end{cases} Define the function T(x)=(f(x)−g(x)) g(x)f(x) g(x)−1.T(x)=\dfrac{\bigl(f(x)-g(x)\bigr)\,g(x)}{f(x)\,g(x)-1}. Answer the following. a. Part 1: (i) State the domain of ff. Give your answer using set notation. b. Part 2: (ii) State the domain of gg. c. Part 3: (iii) Determine the set of real numbers for which the numerator (f(x)−g(x)) g(x)(f(x)-g(x))\,g(x) is defined. d. Part 4: (iv) Determine the set of real numbers for which the denominator f(x) g(x)−1f(x)\,g(x)-1 is defined and non-zero. e. Part 5: (v) Hence, obtain the complete domain of TT. Present your answer in set-builder form.
Intuitive explanation

This problem asks you to work with piecewise functions and compose them into a more complex rational expression. The key insight is that T(x)T(x) is only defined where its denominator is nonzero—that is, where f(x)g(x)≠1f(x)g(x) \neq 1—and you'll need to determine which piece of ff and which piece of gg apply in different regions of the domain. Once you identify the relevant pieces for a given xx-value, substitute them into T(x)T(x) and simplify, being careful about domain restrictions that make ff or gg undefined or make the denominator zero.

Intuitive explanation

This problem asks you to work with piecewise functions and compose them into a more complex rational expression. The key insight is that T(x)T(x) is only defined where its denominator is nonzero—that is, where f(x)g(x)≠1f(x)g(x) \neq 1—and you'll need to determine which piece of ff and which piece of gg apply in different regions of the domain. Once you identify the relevant pieces for a given xx-value, substitute them into T(x)T(x) and simplify, being careful about domain restrictions that make ff or gg undefined or make the denominator zero.

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