I am completely unsure whether the following are "equations" and therefore I'm unsure of where to begin solving them. 1. The problem statement, all variables and given/known data In exercises 39-46, f refers to the function with domain [0,2] and range [0,1], whose graph is shown in Figure P.55 (how do I draw and paste graphs? the graph looks like an upside down curve with a maximum height or range of (1,1) and a width or domain of (0,2)). Sketch the graph of the indicated functions and specify their domain and ranges. 2. Relevant equations 39. f(x) + 2 41. f(x + 2) 43. -f(x) 45. f(4 - x) 3. The attempt at a solution No idea.
It can help, although mathematicians might cringe at the idea, to consider the domain of a function as its "input", the range is the "output". In the equations you posted, x varies over the domain, f "takes" x to the range of f. Domain [0,2] means the interval [0,2] along the x axis, range [0,1] is the y axis--the domain and range are perpendicular intervals.
Not true Emil, in this case it doesn't matter it's obviously implied. y = f(x) y = f(x) + 2 Will shift every value up by 2 y = f(x + 2) Will shift every value right by 2 y = -f(x) Will reflect it about the y axis. y = f(4 - x) Will reflect it about the x axis and shift it by 4.
ughhh....Chipz, "An equation is a mathematical statement that asserts the equality of two expressions.[1] Equations consist of the expressions that have to be equal on opposite sides of an equal sign"
Im sorry but mathematics is the most rigorous science.Do not assume anything. And in high school by mistake when I wrote an equation without an equals sign,I got the minimum possible note. But what have we here? equations? functions? graphs?So I stopped at the first mistake! You encourage these mistakes? Please Register or Log in to view the hidden image!
I work with scientists, physicists and engineers all day; I understand the requisite rigor in mathematics. I also remember being in elementary / high school and teachers would confuse me with pedantry which was of no assistance to my learning and did nothing but confuse the issue. They were poor teachers. If you want to ask inane questions on content they likely copied directly from their text book (and is thus the fault of the text book printers) then by all means, be a dick. A function is an equation, a graph is a representation of a function. I somehow doubt you got beyond high-school mathematics.
Despite some rather bad-tempered exchanges here, let's see if I can help. First note that for any function \(f\) on sets, the bit in parentheses is called its argument and is, by definition an element in the domain. Whatever happens "outside" the argument, as it were, is immaterial to the function. So to take a couple of examples from your problem set: \(f:[0,2] \to [0,1]\), So that for any \(x \in [0,2]\) then \(f(x) = \frac{x}{2} \in [0,1]\) and this is not altered by the fact that we add a number in the codomain: the result is not in the range of your function. Conversely (not a good choice of words) for the function \(f(x+2)\) the constant is an element in the domain whose image is in the specified range, so you must find a function \(f:[0,2] \to [0,1]\) that satisfies this requirement. Can you do this? Hint: the variable \(x\) alone is NOT specified as an element in the domain; it may "live" anywhere it chooses. Try \( x \in [-?, -?]\) (query marks intended) and see if you can construct this function
The statement and the data requirements of this problem is very imprecise and vague If it is required, then there are a number of n solutions:\(f (x) =-\frac{x }{2} +1 \), or \( f(x)=\ x^2 -2x +1\), or \(f (x) = -\ x ^ 2 +2 x \) etc...
