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does y equal -3 describe a function or do I also need the domain?

does y equal -3 describe a function? or do I also need the domain. to describe a function?

Research, Knowledge and Information :

Functions: Domain and Range | Purplemath

State the domain and range of the following relation. Is the relation a function?
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What is a Function

What is a Function? ... And we usually see what a function does with the input: f(x) ... Also, notice that: the domain is {2,3,7} (the input values)
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1.3 - Functions - Richland Community College

Also, g is a function of both x and y. ... does not equal 3 * f(x) = 3 ... These aren't in the implied domain, so they need to be stated.
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CHAPTER 12 Functions -

Functions Y ... Definition 12.3 Two functions f : A !B and g D are equal ... 1 frombothsidesandinvertingproducesx˘ y. Therefore f isinjective. Function f ...
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Domain of a function - Wikipedia

... the set of values the function takes on as output is termed the image of the function, which is sometimes also ... or equal to 0 (ignoring ... domain of a ...
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6 Ways to Find the Domain of a Function - wikiHow

The domain of a function is the set of numbers that ... Here are the basics that you need to know about each type of function, ... so the domain is equal to all ...
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Domain and Range - Free Math Help

Learn what the domain and range mean, ... The domain of a function is the set of all possible input values, ... We need a function that, ...
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Finding the Domain of a Function |

Can we still find the domain and range? ... (you usually need the ... Asking for the domain of a function is the same as asking
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CorrectionKey=TX-A CorrectionKey=TX-B LESSON 1 . 1 DO NOT ...

that x is greater than or equal to 1.” For interval notation, do the ... Also describe the end behavior ... describe the function’s domain and range using one of ...
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Functions: The domain and range - The University of Sydney

Functions: The domain and range ... we often describe a function using the rule, y = f ... X → Y the domain of f is the set X. This also corresponds to the set of x ...
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Suggested Questions And Answer :

does y equal -3 describe a function or do I also need the domain?

y=-3 is legal funk even tho its konstant & thur Not a funk av x
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How do I graph (-x^2 + 3x +9)/(x-1)

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f(x)=-2x^2+2x+2 ?

need to find domain for f(x) = 2/x^2
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how do you find the domain of the composite function when the f(x)=sqrt of x and g(x)=2x-5

When we look at each function we can see that g is defined for all values of x as a linear function. It is represented by an infinitely long straight line. But f is the square root of g(x) when the two functions are combined. We can't find the (real) square root of a negative number so the domain must be limited. What we need to find out is: when is g(x) negative? To answer this we put g(x)=2x-5<0 and solve for x. 2x<5, so x<2.5. These are values of x we need to avoid. The domain for the combined function f of g is therefore x>2.5. The context of your question has g(x)=2x+5, so g(x)=2x+5<0 when x<-2.5 and the domain of x needs to avoid these values, so x>-2.5. Check which is correct: the title for your question or the context.
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given the folowing piecewise function:

A. The domain is the range of values of x for which the function f(x) is defined. The lowest value for x is -4. There is no upper limit for x, because the function is defined for x>1. Therefore, the domain is x=>-4. B. The range of the function requires us to calculate its value over the domain. For -4<=x<1, the range is 1<=f(x)<6. At x=1, f(x)=9, a singularity. For x>1, f(x)>2 and is unbounded, i.e., goes off to minus infinity as x increases to infinity. C. The intercepts are where the lines cross the axes. We need to know what value of x makes f(x) zero, and what value of f(x) makes x zero. When x=0, f(x)=5; f(x)=0 when x=3. So the intercepts are (0,5) and (3,0).  D. f(x) is not continuous because when x is just less than 1 f(x) is just less than 6, but it jumps to 9 when x is exactly 1, for that single point alone. After that, when x>1 f(x) goes to 2 from 9. E. The graph consists of two lines and a singularity. For -4<=x<1, f(x) goes from 1 to just short of 6 in a straight line. The line joins the points (-4,1) and (1,6), cutting the f(x) vertical axis at (0,5). At exactly x=1 we have the single point (1,9), and after that we have a line starting at point (1,2) and cutting the x-axis at (3,0) and going on indefinitely. The first line has a positive slope (/) and the second line a negative slope (\).  I hope this information is clear enough for you to draw the graph.
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f (x)=(x+e^x)/(x-e^x) find the zeros and domain of this function

GGiven f (x)=(x+e^x)/(x-e^x) zeros of the function (x+e^x)/(x-e^x)=0 (x+e^x)=0 x=-0.567143 Domain of the function is all real R Precalculus Help -
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How are the locations of vertical asymptotes and holes different, and what role do limits play?

To talk about asymptotes and holes, you need pictures. These pictures are graphs of functions. The simplest function containing vertical and horizontal asymptotes is y=1/x, where x is the horizontal axis and y the vertical axis. The vertical asymptote is in fact the y axis, because the graph has no values that would quite plot onto the y axis, although the curve for 1/x gets very, very close. The reason is that the y axis represents x=0, and you can't evaluate 1/x when x=0. You' d have to extend the y axis to infinity both positively and negatively. You can see this if you put a small positive or negative value for x into the function. If x=1/100 or 0.01, y becomes 100. If x=-1/100 or -0.01, y becomes -100. If the magnitude of x decreases further y increases further. That's the vertical asymptote. It represents the inachievable. What about the horizontal asymptote? The same graph has a horizontal asymptote. As x gets larger and larger in magnitude, positively or negatively, the fraction 1/x gets smaller and smaller. This means that the curve gets closer and closer to the x axis, but can never quite touch it. So, like the y axis, the axis extends to infinity at both ends. What does the graph look like? Take two pieces of thick wire that can be bent. The graph comes in two pieces. Bend each piece of wire into a right angle like an L. Because the wire is thick it won't bend into a sharp right angle but will form a curved angle. Bend the arms of the L out a bit more so that they diverge a little. Your two pieces of wire represent the curve(s) of the function. The two axes divide your paper into four squares. Put one wire into the top right square and the other into the bottom left and you get a picture of the graph, but make sure neither piece of wire actually touches either axis, because both axes are asymptotes. The horizontal axis represents the value of x needed to make y zero, the inverse function x=1/y. Hence the symmetry of the graph. Any function in which an expression involving a variable is in the denominator of a fraction potentially generates a vertical asymptote if that expression can ever be zero. If the same expression can become very large for large magnitude values of the variable, potentially we would have horizontal asymptotes. I use the word "potentially" because there's also the possibility of holes under special circumstances. Asymptotes and holes are both no-go zones, but holes represent singularities and they're different from asymptotes. Take the function y=x/x. It's a very trivial example but it should illustrate what a hole is. Like 1/x, we can't evaluate when x=0. However, you might think you can just say y=1 for all values of x, since x divides into x, cancelling out the fraction. That's a horizontal line passing through the y axis at y=1. Yes, it is such a line except where x=0. We mustn't forget the original function x/x. So where the line crosses the y axis there's a hole, a very tiny hole with no dimensions, a singularity. So a hole can occur when the numerator and denominator contain a common factor. If this common factor can be zero for a particular value of x, then a hole is inevitable. Effectively it's an example of the graphical result of dividing zero by zero. With functions we can't simply cancel common factors as we normally do in arithmetic. Asymptotes and holes are examples of limits. Asymptotes can show where functions converge to a particular value without ever reaching it. Asymptotes can be slanted, they don't have to be horizontal or vertical, and they can be displaced from both axes. Graphs can aid in the solution of mathematical and physics problems and can reveal where limitations and limits exist for complicated and complex functions. Knowing where the limits are by inspection of functions also aids in drawing the graph. This helps in problems where the student may be asked to draw a graph to show the key features without plotting it formally.
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find domain,and locate intercepts; f(x)=x+3 f(x)=-2x-3 if x<-2 if x>=-2

I interpret this as f(x)=x+3 when x<-2 and f(x)=-2x-3 when x>-2, because the whole domain of x is covered by one or other function. Treating the functions separately, the domain of the first function is all x<-2; the domain of the second function is all x>-2. When x=0 we have the y or f(x) intercept and the second function applies, so the y intercept is -3. When y=f(x)=0 the x intercepts are -3 when the first function applies (-3<-2), and -3/2 when the second function applies. (-2)=1 and when x is just less than -2 f(x) approaches 1. The slope of f(x) is positive (/) when x<-2 and negative (\) when x>-2. Therefore the graph represents an inverted V with apex at (-2,1) and it cuts the x axis at -3 and -3/2. The left part of the graph can be drawn by joining (-2,1) to (-3,0) and extending the line below the x axis where x<-3 and y<0; and the right part can be drawn by joining (-2,1) to (0,-3), cutting through the x axis at (-3/2,0) and extending the line below the x axis where x>0 and y continuing negative. The range is f(x)<1.
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how do you write an equation for the linear function f with the given values?

Standard linear function is f(x)=ax+b where a and b are constants. Substitute each point into the function: 21=-2a+b, -35=5a+b. We need go no further because we have two linear equations and two unknowns. f(-2)-f(5): 56=-7a so a=-8. Now we can work out b: b=21+2a=21-16=5 so f(x)=5-8x. Quick check shows that the function is correct: f(-2)=5+16=21; f(5)=5-40=-35. But we need to check all the other points: f(-6)=5+48=53; f(3)=5-24=-19; uh-oh, something wrong! The values don't fit. So either f(x) is not linear or it's piecewise. Closer inspection is needed. First plot the points. It's clear to see that the points are not colinear. Join the points with straight lines. We're not told that f is continuous. Let's assume it is. If it's piecewise, we need 5 different equations to define the function between the 6 points. In order these are (-9,-4), (-6,-2), (-2,21), (3,-5), (5,-35), (12,14). Call these points A, B, C, D, E, F. We can work out linear equations between A and B, B and C, C and D, etc. These equations will provide continuity for f in the domain -9 Read More: ...

what is a piecewise graph?

​A piecewise graph is one made up of more than one function definition. For a particular domain of the x value,  between certain values, or bigger or smaller than a particular value one function applies, while for a different domain another function applies, and so on for as many function definitions as you're given. Usually the different domains meet so as to provide a continuous domain for x. They can't overlap.
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