📝 Composition of Functions (17 MCQs)

📖 Explanation: The composition \(f\\circ g)\ means we first apply \g\ to \x\ and then apply \f\ to the result, giving the formula \(f\\circ g)(x)=f(g(x))\. This is the standard definition used throughout calculus.

📖 Explanation: For the composition to be defined, \x\ must first belong to the domain of \g\; after evaluating \g(x)\, that value must lie in the domain of \f\. Hence the correct condition is that \x\ is in the domain of \g\ and \g(x)\ is in the domain of \f\.

📖 Explanation: To have \\\sqrt{g(x)}\ defined we need \g(x)\\ge 0\; that is \x-2\\ge0\ which gives \x\\ge2\. There are no other restrictions, so the domain is the interval \[2,\\infty)\.

📖 Explanation: Since \(f\\circ g)(2)=f(g(2))\, we have \f(1)=5\. The given values directly determine the needed output of \f\ at the argument \1\.

📖 Explanation: A decreasing function \g\ reverses the order of inputs; applying an increasing function \f\ to those outputs preserves the reversal, so the overall composition \f\\circ g\ is decreasing on the interval.

📖 Explanation: The equation \(f\\circ g)(x)=x\ means that applying \g\ then \f\ returns the original \x\; therefore \g\ acts as a right inverse of \f\ (i.e., \f\∘\g\=Id).

📖 Explanation: Evaluating each piece shows that at \x=0\ we have \g(0)=-0=0\ and \f(0)=-1\; both compositions give \-1\. At \x=1\, \g(1)=-1\ and \f(1)=0\; again both yield \0\. No other \x\ satisfy the equality, so the set is \\{0,1\\}.

📖 Explanation: Compute \f(g(x))=(x+3)^{2}=x^{2}+6x+9\ and \g(f(x))=x^{2}+3\. Equating gives \6x+9=3\ so \6x=-6\ and \x=-1\. Thus \-1\ is the unique solution.

📖 Explanation: For \f\\circ g\ we need \x\\ge0\ for \\\sqrt{x}\ and \\\sqrt{x}\\neq1\ (i.e., \x\\neq1\). Hence \[0,\\infty)\\setminus\\{1\\}\. For \g\\circ f\ we require \x\\neq1\ and \1/(x-1)\\ge0\, which holds when \x>1\. Thus the domain is \(1,\\infty)\.

📖 Explanation: Associativity follows from the definition: \f\\circ(g\\circ h)(x)=f(g(h(x)))\ and \(f\\circ g)\\circ h)(x)=f(g(h(x)))\. Since both expressions are identical for any \x\ where the compositions are defined, the property holds whenever the individual compositions are defined.

📖 Explanation: If \y=(f\\circ g)(x)=f(g(x))\, applying \g^{-1}\ first gives \g^{-1}(f^{-1}(y))\. Hence the inverse of the composition is \g^{-1}\\circ f^{-1}\. This holds for any bijections, not just linear ones.

📖 Explanation: First evaluate \h(2)=2^{3}=8\. Then \g(8)=1/8\. Finally \f(1/8)=\\sqrt{1/8}=1/(2\\sqrt{2})\. Hence the composition yields \1/(2\\sqrt{2})\.

📖 Explanation: Applying \g\ first gives \g(x)=\\ln x\ (defined for \x>0\). Then \f\ squares this result, so \p(x)=(\\ln x)^{2}\. The composition inherits the domain of \g\, namely all positive real numbers.

📖 Explanation: Take \g:\\mathbb{R}\\to(0,\\infty)\ defined by \g(x)=e^{x}\ and \f:(0,\\infty)\\to\\mathbb{R}\ defined by \f(y)=\\ln y\. Then \(f\\circ g)(x)=\\ln(e^{x})=x\ for every real \x\, while \(g\\circ f)(y)=e^{\\ln y}=y\ only for \y>0\; thus \g\\circ f\ is not the identity on all of \\\mathbb{R}\.

📖 Explanation: A simple counterexample, such as \f(x)=x^{2}\ and \g(x)=x+1\, gives \(f\\circ g)(x)=(x+1)^{2}\ and \(g\\circ f)(x)=x^{2}+1\, which are not equal. Hence composition does not generally satisfy the commutative property.

📖 Explanation: Since \g\ requires \x>0\, \(f\\circ g)(x)=e^{\\ln x}=x\ for \x>0\. The composition \g\\circ f\ uses \f(x)=e^{x}>0\ for any real \x\, so \(g\\circ f)(x)=\\ln(e^{x})=x\ with domain \\\mathbb{R}\.

📖 Explanation: The composition of two bijections is itself a bijection. Its inverse reverses the order of application: first apply \f^{-1}\ then \g^{-1}\. Hence \h^{-1}=g^{-1}\\circ f^{-1}\. This follows directly from the definition of inverses for one-to-one and onto functions.