🔬 Problem set gamma#

\(\gamma\).1#

Let \(A\),\(B\) and \(C\) be any three sets. Show that \(A \cap (B \cup C) = (A\cap B) \cup (A \cap C)\).

One way to show that \(E=F\) is show that a arbitrary element of \(E\) must also be in \(F\) and vice versa.

\(\gamma\).2#

Let \(A\), \(B\), \(C\) and \(D\) be some set such that \(A \subset C\) and \(B \subset C\). Let \(f\colon D \rightarrow C\) be a function.

Show that \(f^{-1}(A \setminus B) = f^{-1}(A) \setminus f^{-1}(B)\).

\(\gamma\).3#

Find the composition \(g \circ f\) of two functions \(f\) and \(g\), if it exists:

  1. \(f \colon \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)=\sin(x)\) and \(g \colon \mathbb{R} \rightarrow \mathbb{R}\) defined by \(g(x)= \frac{x}{1+x^2}\)

  2. \(f \colon \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)= 1-x^4\) and \(g \colon (1,\infty) \rightarrow \mathbb{R}\) defined by \(g(x)= \log(x-1)\)

  3. \(f \colon \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)=\cos(x)\) and \(g \colon \mathbb{R}\setminus\{1\} \rightarrow \mathbb{R}\) defined by \(g(x)= \frac{x}{1-x}\)

Is there a composition in each case?

\(\gamma\).4#

Let \(f\) and \(g\) be any two functions from \(\mathbb{R}\) to \(\mathbb{R}\). Is it true that \(g \circ f = f \circ g\) always holds?

There are two things implicit in this question.

First, there is an implicit final sentence here, which is: If yes, prove it. If no, give a counterexample.

Second, an equality sign between two functions means that they are the same function. Hence to show equality you need to show that they agree everywhere on the domain. To show inequality, you need to give just one point in the domain where the function values differ.

\(\gamma\).5#

For each of the following functions, what is its domain and range? Is it one-to-one? Is it onto? Is it bijective? If it is bijective, write the expression for the inverse function.

  1. \(f(x) = 3x-7\)

  2. \(f(x) = e^x\)

  3. \(f(x) = \ln(x)\)

  4. \(f(x) = \sqrt{x-1}\)

  5. \(f(x) = x^2-1\)

Draw a graph for each function.