Title: Functions
Course: Demonstrating STACK (v4.4.7)

For the following function, what is the domain determined by the domain convention? \[ f([value]) = \sqrt{[value]^2-[value]} .\] Give your answer as a logical sequence of inequalities such as "\( x<= -7\mbox{ and } x>5\)".
Feedback: The function is only defined where the square root exists, i.e. where \[ [value]^2-[value]^2 >= 0.\] This is true for \((-\infty, -[value]] \cup [[value], \infty).\) This is entered as \([value] \le -[value]\) OR \([value]\ge [value]\).

For the following function, what is the domain determined by the domain convention?

\[ f([value]) = \frac{1}{\sqrt{[value]-[value] [value] -[value]^2}} .\] Give your answer as a logical sequence of inequalities such as "\( x<= -11 \mbox{ or } x>13\)".
Feedback: For the function to exist, the square root must exist and cannot be \(0\), hence \[[value]-[value] [value] -[value]^2 = -([value] + [value])([value] - [value]) > 0.\] This is the case when \[ -[value] < [value] < [value]\] This is entered as \([value] \gt -[value]\) and \([value]\lt [value]\). Enter \(-[value] < [value] \mbox{ and } [value] < [value].\)

A function \( f : R \rightarrow R\) (where \(R\) denotes the real numbers) is defined by the rule\[ f(x) = [value]\] What is the output of the function \(f\), when the input is [value] ?
Feedback: You need to evaluate the function at the value of \(x\), i.e. \(f([value]) = 9^{([value])}= [value].\)

A function \( f : R \rightarrow R\) is defined by the rule \[ f(x) = [value].\] Which input of the function \(f\), has the output [value]?
Feedback: To find the input, we need to find the value of \(x\) for which \[[value] = [value].\] It can easily be checked that this is true for \(x = [value]\).

Find the Fundamental period of the following periodic function. \[ f(x) = [value] \cos\left([value] \pi x \right). \]
The Fundamental Period =
Feedback: The Fundamental Period of \(\cos(x)\) is \(2 \pi\), so we need a number \(P\) such that \[[value]\pi(x+P) = [value]\pi x + 2\pi,\] which is true when \(P = [value].\)

Determine the image set \({\rm im}(f)\) of the function\[ [value].\] Give your answer as an inequality in \(y\), e.g. \( y <= \alpha\).
Feedback: The function can be rewritten by completing the square as \[[value]=(x+[value])^2+([value])\] so that all values are larger than \([value]\), and for \(x=-[value]\) equals \([value]\). Hence \[ {\rm im}(f) = \{y| y >= [value]\}\] and the answer is \[ y >= [value].\]