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Domain
Find the domain of y = x + 4
Domain simply means "what numbers can I use for X that give me an answer for Y?"
Domain is looking for the acceptable X values that work in the equation y = x +4
When looking for the domain, the rule of thumb is:
* No fraction? * No radical? (sqrt, cube root, etc) * No problems! If you don't see a fraction, or a radical, or both in your equation, the domain will always be "All real numbers". This means you can pick any number you want, plug it in for X, and you'll get an answer for Y.
If you do happen to have a fraction, radical (or both)...
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Think about what "breaks" a fraction. What makes a fraction not work?
If you try to divide by 0, you get an error, or undefined on your calculator.
for example:
Find the domain of y = 1/(x -1)
1/x is a fraction, and we know having a 0 on the bottom would make it not work.
So what we do is say the domain is "all real numbers, except" and then find out what numbers break the function and fill that in later.
We set the bottom of the fraction = 0 to find out exactly which numbers will break it, and then solve for X.
x - 1= 0
x = 1
So we know that is we plug in a 1, it will break the fraction.
The domain of y = 1/(x - 1) is "All real numbers, except 1"
---- If you have a radical expression in your function:
example. y = sqrt(x+2)
We need to know what breaks a radical expression, what number(s) won't work?
Well, sqrt(0) is OK. it = 0.
sqrt(1) is ok, it = 1.
But what happens if I try sqrt(-1). Try it on your calculator.
You get an error message, right?
This is because you cannot take the root of a negative number. (at least, not yet)
This means, negative numbers break a function with a radical. So similarly to how we found the numbers that broke a fraction, we'll set what's inside the radical less than 0. Since negative numbers are less than 0.
x + 2 < 0 and solve for x
x < -2
So any number less than -2 will break my radical.
Your answer would be, "All real numbers >= -2" (since anything less than -2 is broken, but -2 is still OK)
To check:
y = sqrt( -2 + 2) = sqrt (0) = 0 - OK
y = sqrt( -3 + 2) = sqrt(-1) - breaks
I chose a number smaller than -2 to check.
---- If you get a problem where they use both fractions AND radicals, just use both techniques.
Ex: y = 1/sqrt(x + 2)
We know that having a 0 on the bottom of a fraction breaks it, but it OK for a radical to have a 0 in it. We need to combine both rules together.
Take the inside of the radical and set it less than(what breaks the radical) or equal to 0 (what breaks the fraction).
x + 2 <= 0
x <= -2
This means, any number less than or equal to -2 will break both the radical and the fraction.
This tells us that the domain has to be "All real numbers > -2" (not including -2 this time)
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Find domain y equals x-4?
The domain would be: all reall numbers. The domain is which 'X' values you can insert into 'X' and get a value for 'Y'. Since any real number could be inserted to create a 'Y' value, the answer is:all real numbers.
What is the domain range asymptote and intercept of the equation y equals 4 times 2 exponent x?
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
What is the Domain and Range of y equals the square root of x minus 4?
In the complex field, the domain and range are both the whole of the complex field.If restricted to real numbers, the domain is x >= 4 and y can be all real numbers >= 0 or all real numbers <= 0 [or some zigzagging pattern of that set].
What is the domain of the function y sin x?
y= sin 3x
Determine the inverse g(x) of the function f(x) = (1+(4/x)), stating its domain and range. Verify that f(g(x)) = g(f(x))=x and that gā²(f(x)) = (1/(fā²(x))?
Let f(x) = y y = 1 + (4/x) Now replace y with x and x with y and find equation for y x = 1 + (4/y) (x-1) = (4/y) y = 4/(x-1) This g(x), the inverse of f(x) g(x)= 4/(x-1) The domain will be all real numbers except when (x-1)=0 or x=1 So Domain = (-ā,1),(1,+ā) And Range = (-ā,0),(0,+ā) f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x So we get f(g(x)) = g(f(x)) Notice the error in copying the next part of your question It should be g'(f(x)) = 1/(f'(g(x))) g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1 f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1 1/[f'(g(x))] = 1/1 = 1 g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)
What is the Domain and range of x 2 plus y equals 4?
To find the domain or range, solve for a variable and see if the other variable has any restrictions on it. In this case, x2 + y = 4 y = 4 - x2 There are no restrictions on x, therefore x is in the domain of all real numbers. x = square root(4 - y) Since the argument (number in brackets) of a square root must be positive, 4 - y > 0, y < 4. Domain: x can be all real numbers. Range: y can be all real numbers less than or equal to 4.
Is the domain the y or x?
The domain is the x or the input
Find domain y equals x-4?
The domain would be: all reall numbers. The domain is which 'X' values you can insert into 'X' and get a value for 'Y'. Since any real number could be inserted to create a 'Y' value, the answer is:all real numbers.
What is the domain range asymptote and intercept of the equation y equals 4 times 2 exponent x?
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
Is the domain the x or y value?
The domain is the x value
What is the domain and range of y equals the square root of x squared plus 4?
(x^2)^(1/2) equals x, therefore, y = x+4, which has a range and domain of all real numbers. The graph is a straight line, slope of 1, y-intercept of 4. Are you actually saying y = (x^2+4)^(1/2). If so, the range and domain will also be all real numbers because x^2+4 will never result in a negative number.
What is domain in mathematics?
domain = x-values range = y-values for which x or y is a solution
What is the range for y equals x plus 4?
It depends on the domain but, if the domain is the real numbers, so is the range.
Find the domain and range of f of x equals x squared minus 4?
D = {x [element of reals]}R = {y [element of reals]|y >= 4}
Is the domain the x or the y axis?
The domain of a relation is the X axis.
Is the domain dependent or independent?
domain is independent why? because its before range or also known as x/domain and y/range(x,y).
What is the domain of y equals x2-x-2?
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