Find f '(x), and use interval notation to give the domain of f.?
f(x) = x^2ln(8-4x^2) i know that i have to use the product rule, but how do you find the derivative of the second part? i tried 2x/8-4x^2, but that's wrong. i know that if it were 8-4x, it would be 4/8-4x, but the squared changes the numerator. also, the question asks for the domain of f. is that the domain of f(x) or f '(x)? Thanks a lot.
Graph the function g(x) = 3^x-1 and give its domain and range using interval notation?
http://www.wolframalpha.com/input/?i=3^x... Domain: (-infinity, infinity) Range: (-1, infinity)
What is the most intuitive way prove the derivative addition rule?
Intuition does not make for good proofs. At best, it can show you why something “should” be correct, but sometimes it fails to account for all the ways that something could occur, or otherwise fail to be complete. If you want to see why it makes sense for something in math to be true, intuition can help; but always go for a formal proof to be sure.Intuitively, the derivative is a small change in the output caused by a small change in the input. If you have two functions added together, the small change of their sum should be the sum of their individual small changes.But in this case, why rely on intuition? John Conway (famous mathematician) likes the concept he calls “one-line proofs” — a proof which consists of a bunch of statements or expressions connected by “if and only if” connectives or equality. In theory, the entire proof is one statement of the form “[math] A \iff B \iff C \iff D[/math]” (or [math]A = B = C = D[/math]. They don’t require deep thought, just a simple series of obvious equalities/equivalences connecting the beginning and end.The addition rule for derivatives has such a one-line proof:[math]\frac{d}{dx}(f+g)(x) = \lim_{h\to0}\frac{(f+g)(x+h)-(f+g)(x)}{h} = \lim_{h\to0}\frac{(f(x+h)+g(x+h))-(f(x)+g(x)}{h} = \lim_{h\to0}\frac{(f(x+h)-f(x))+(g(x+h)-g(x)}{h} = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} + \lim_{h\to0}\frac{g(x+h)-g(x)}{h} = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)[/math]Each equality follows either from the definition of the derivative or a simple rearrangement of additive terms. One equality does use the additive property of limits ([math]\lim (f+g) = \lim f + \lim g[/math]), but that should be known by the time you start dealing with differentiation, you should already know the rules for limits.
Find domain equation, graph function interval notation, (f-g)(3)?
1) f(x) is well defined except where 4x²-5x+1 = 0 => (4x-1)(x-1)=0 => x = 1/4 or x = 1 So the domain is the set of real numbers except 1/4 and 1. You can write that in set notation as (-∞,1/4) ∪ (1/4,1) ∪ (1,∞) 2) If y = (x+5)/2 then x+5 = 2y x = 2y-5 so to get x from y we need f^(-1)(y) = 2y-5 and changing the names of the variable f^(-1)(x) = 2x-5 3) f(x) is well-defined for all real numbers. So the domain in interval notation is (-∞,+∞) 4) Not sure what you mean here. Do you really mean f-g? Or is it composition of functions?
How do you solve (f-g)(x) using the given graph?
This is a question about subtraction and how it is defined for functions. Simply put:[math]\left ( f-g \right )\left ( x \right ) = f\left ( x \right ) - g\left ( x \right )[/math]e.g. [math]f\left ( x \right ) = 2x[/math][math]g\left ( x \right ) = 7x[/math][math]\left ( f-g \right )\left ( x \right ) = 2x - 7x = -5x[/math]The complication in this particular question is we don't have an explicit formula for either f(x) or g(x), so precise numbers would be impossible to get. Good thing these questions don't ask for them.a. We can tell to a practical extent when the function (f - g)(x) is positive by looking at the provided graph, as it will tell us the value of f(x) or g(x) at any x from -5 to 5. For instance, [math]\textrm{when }\;x = 0,\; f\left ( x \right ) = 1 \; \textrm{ and } \; g\left ( x \right ) = 0[/math]. Thus, [math]\left ( f-g \right )\left ( 0 \right ) = f\left ( 0 \right ) - g\left ( 0 \right )= 1-0 = 1[/math]. Clearly, (f - g)(0) is positive. Note that with the graph, we can tell when the function (f - g)(x) is positive or negative by comparing the relative positions of the graphs of f(x) and g(x). The graph of f(x) is above the graph of g(x) from -5 to approximately 3; therefore, the function (f - g)(x) is positive on the interval [-5, 3].*b. In a similar vein, when the graph of f(x) meets the graph of g(x), the function (f - g)(x) = 0, and when the graph of f(x) is below the graph of g(x), the function (f - g)(x) returns a negative output.c. To sketch a graph of (f - g)(x), you'll have to approximate the difference between f(x) and g(x) at some values of x, then graph those differences on the y-axis at the respective points on the x-axis.*I suppose your teacher expects you to to answer with the interval (-∞, 3], but the weakness of the graph-only approach is that we really can't be sure how the function actually behaves. It could go straight up after -5 for all we know.
What are some tips to help a beginner start improvising on the piano?
When I started piano lessons, I was forbidden to improvise or play by ear. I was supposed to play only what was on the printed page. My teacher was concerned that I learn fluency in reading music, and I followed directions for a couple of years. Being an obedient student, however, inhibited me and made it more difficult for me to see a piece as a whole rather than an image stave by stave. Once I did begin to improvise on my own, I learned a great deal about piano technique and found emotional satisfaction in doing so. I was able to choose what chords and scales, and what bizarre, atonal patterns I could create. You didn't mention what style you're most interested in--classical, pop or jazz--but in all of them you will gain skill in your fingers by spending time in free improvisation. This is an excellent way to make basic chord patterns and the ways of arpeggiating them second nature to your hands.A good start is to take a piece you like and examine the principal chord patterns, whether it's a basic I-IV-V-I or Mozartean [spell check thought that should be "smog artisan"--not a classical music lover!] I-VI-IV-II-V-I or something more remote, and play that apart from the song in different ways. You can break up the chords in different ways, you can play it in different keys, you can substitute a new chord for a chord that's already there and see if it's pleasing or a dud.However you go about it, playing the elements of music in ways you find interesting will also help you become a better musician. One of the great things about playing piano is that you are "the master of your domain," which gives you the freedom to go at your own pace and learn what is satisfying to you.Debussy would never have become the original composer he did had he not spent his childhood searching out chords and timbres at the piano which intrigued and pleased him. Now, get started enjoying your musical future!
How do I know whether an inverse of a function is differential at a certain value?
The inverse of f is not differentiable at a point where the derivative of f is 0. Find these points and check if f=-8 at that point because the y value of f becomes the x value of the inverse function.So at c=-8 the derivative of the inverse is undefined. You can even sketch the graph to know this.Vertical tangent line means infinite gradient, which means derivative is undefined.
I'm told that two functions are equivalent if their outputs are equal on the intersection of their domains and they are identical if their domains and codomains are the same. Is this an official definition?
No. Formally, a function [math]f:A\rightarrow B[/math] is a subset of [math]A\times B[/math], so that each element of [math]A[/math] occurs exactly once as the first part of an element in [math]f[/math]. [math]f(x)[/math] is then notation for "the second part of the unique element in [math]f[/math] for which [math]x[/math] is the first part".Thus, equality of functions is inherited from equality of sets: two functions [math]f,g[/math] are equal if their domain and codomain are equal, and if the image of both functions is the same for each element in the domain."Equivalence" is not canonically defined for functions, because several notions exist. Generally it is safe to say two functions are equivalent if the domain and the images are the same, but not necessarily the codomain (I might even go so far as to call them "equal", since they are for most if not all intents and purposes). Other measures of "equivalence" could depend on the context, for example, two functions are considered equivalent in Fourier analysis if they are the same in almost all points (that is, all but a countable number) of their codomains.(What follows is non-canon, but are reasonable definitions)What you propose is a good definitions of saying that two functions are equivalent on a certain subset of their domains. That is, given a set [math]A[/math], two functions [math]f,g[/math] whose domain is a superset of [math]A[/math] can be called equivalent if for each [math]a\in A:f(a)=g(a)[/math]. This equivalence could be denoted [math]f\sim_A g[/math].This is not a good notion for general equivalence of two functions, because we need to fix the set [math]A[/math] first. Otherwise, it is not an equivalence relation, being non-transitive.I would call the relation you propose, which is in the terminology above "f and g are equivalent on the intersection of their domains", compatibility.
What are some interesting examples of uncountable sets?
Here's an uncountably infinite set that I found to be interesting. It is the set of functions whose domain and codomain are the natural numbers, N = { 1, 2, 3, ... } Functions like this can be expressed using the notation f: N -> N.To be such a function, it must be defined for every natural number, and the result must be a natural number.Here are some examples:Let f: N -> N where f(n) = 2n + 3. Some values are f(6) = 15, and f(4) = 11.Let f: N -> N where f(n) = n^2. Some values are f(3) = 9, and f(5) = 25. Let f: N -> N where f(n) = the n'th digit of pi. The first few values are f(1) = 3, f(2) = 1, and f(3) = 4.It might not seem like there are an uncountably infinite number of such functions, but there are.Here are some interesting results that follow from this fact.The set of Java programs (or the set of programs in any language) is countably infinite, because you can list them according to the number of characters they contain. But since there are an uncountably infinite number of functions from N to N, then there are some functions which cannot be written in Java, or any programming language. Essentially, this means there are some functions which cannot be computed!Not to be confusing, but there are an uncountably infinite number of functions from N to N that cannot be computed, so there are 'more' functions that cannot be computed than ones that can be.-----If you want to get a bit more advanced here, let me give you a well-defined function that cannot be computed. List the Java programs in order of size. Number these programs 1, 2, 3, etc. Now define the function, f, so that:f(n) = 1 if the n'th Java program halts on all legal inputs, andf(n) = 0 if the n'th Java program does not halt on all legal inputs.So if, for example, f(13) = 1 , this would mean the 13th Java program would halt on all legal inputs.There are many values of this function that we can assign the correct value of 0 or 1 to.But Alan Turing proved in 1936, that it was impossible to compute this function.To be more specific, we'd like to write a Java function that looks like this:int haltsOnAllInputs( int n ){.....}where n if the n'th Java program, and where this Java method returns 0 if the n'th Java program does not halt on all inputs, and returns 1 if it does.Turing showed this is not possible to do.There you have it - a well-defined function that cannot be computed. Go figure.