Q&A for How to Find the Domain of a Function

If your function is a fraction, set the denominator equal to 0 and solve. The domain would then be all real numbers except for whatever input makes your denominator equal to 0. For a square root, set whatever is inside the radical to greater than or equal to 0 and solve, since you can’t use any inputs that produce an imaginary number (i.e., the square root of a negative).

Yes. You can use a graphing calculator to calculate domain by plotting the function. There are also a variety of domain and range calculators online. Simply input your function to find the domain, which is a set of x-values that will successfully generate y-values.

The function tan(90x/2) is undefined at 90x/2 = pi/2 + pi*n, where n is an integer. Simply solve for x to obtain pi/90 + pi*n/45, where n is an integer.

There are no mathematical restrictions on those functions, but the economic context likely imposes an implied domain restriction. Probably x can't be negative or x must also be an integer (one can't sell half cars, for example).

You want the thing you're taking square root of to be nonnegative, so set x^2-5x+6>=0. Solve that for a domain of (-inf, 2] U [3, inf).

This site might have mangled your formatting, so apologies if I'm answering the wrong question. It looks like you're describing a function r which takes real numbers and outputs vectors. If so, then find the domain of each individual component of r. Then the domain of r is the intersection of the domains of each component. If each component is a polynomial like r(t) = 5t+1 or r(t) = t^2, then the domain is all of R.

Since sqrt(x) >= 0, the domain is [0,infinity), or all non-negative numbers.

The domain of y = 3x + 8 is all real numbers, since 3x is a linear function. Linear functions are polynomials, therefore having all real numbers.

You cannot divide by zero, so x^2 - 1000 != 0. Solve to get x = +- 10, so the domain is the real numbers except x = +-10.

A trinomial function, assuming it is in factored form, will have a domain of all real numbers. When you FOIL out the trinomials, you will get a resulting cubic function. Cubic functions are polynomials, therefore having a domain of all real numbers.

Since g is defined in terms of f, you need 2-x to be in the domain of f. The usual trick here is to check the endpoints of the domain, e.g. solve 2-x = -2 => x = 4, and 2-x = 3 => x = -1. Then check what happens on each interval. If x is less than -1, 2-x is too big; x=-1 doesn't work either because the domain of f is open on that side; but between -1 and 4 everything works; x=4 still works because the interval is closed on that side; and x greater than 4 is again outside the valid range. To summarize that in interval notation, write that the domain of g = (-1,4].

Let's assume you meant square root of x + 6 / square root of x - 4. First, x + 6 must be greater than/equal to zero, or x must be greater than/equal to -6. Then, square root of x - 4 means that x - 4 also have to be greater than equal to zero, or x is greater than/equal to 4. Square Root of x - 4 cannot be 0, so x cannot be 4. Then draw a number line and plot these numbers on a number line.

Let's assume the function is (3x - 8) / (x² - 9x + 20). As shown above in Method 2 Step 2, we exclude any x-values that give the denominator the value of 0. (x² - 9x + 20) = (x - 5)(x - 4). Set that expression equal to zero to find that if x = 4 or 5, the denominator becomes 0, and so we exclude those two values of x from the domain. Then the domain is all other values of x.

x can be any real number greater than 0.

The denominator cannot equal 0, because such a denominator would render the function as undefined. So x cannot be 1 or -1. x can be any other real number.

All real numbers.

The domain and range both consist of all real numbers.

All real numbers except 0.

3 is not a function, it is a constant. Therefore, there is no domain.

x can be any real number except 2.

(x² - 9) has to be greater than or equal to 0. Otherwise, we would have imaginary or complex numbers. So x^2 - 9 ≥ 0, x^2 ≥ 9, and x ≥ +/-3, which means that x ≥ 3.

Let's assume the function is y = √(6 - x). (6 - x) must be ≥ 0, so 6 ≥ x. (That's so the function is not the square root of a negative number.)

(x - 2) has to be greater than 0, meaning that x has to be greater than 2. So the domain is all real numbers greater than 2.

The domain is all values of x that don't cause the denominator of that fraction to equal zero (because you can't divide by zero). So let (3x - 7) = 0 and solve for x: 3x = 7, and x = 7/3. That means that the denominator is zero when x = 7/3. So the domain of f(x) is all real values of x except 7/3.

f(x)=5-2x, By definition, f is defined if and only if 5-2x is not equal to Zero. 5-2x=0, -2x=-5, x=5/2, Domain is all real numbers, except for 5/2.

It is the set of all real numbers, since every real number makes sense there.

The domain for this function is a set of all rel numbers. If you plot the graph of this function on a Cartesian plane, you will get function defined on the whole X-axis.

X^2+2^2>0 then x=(-2,2). Then Domain =(-2,2) 【OR】 X^2+4>0,Then subtract -4 on both sides X^2>-4 Then do √ing on both sides ,Then X=+2,-2.then Domain =(2,-2).

There are no restrictions to the value of x, because it is possible to input any x value into this equation and get a y value out. Therefore, the answer is (-inf,inf).

x^2 - 3x + 2 > 0. By Using quadratic equation. -b±√(b^2-4ac)/2a, we get; (-(-3)±√(-3)^2 - 4(1)(2))/2(1)=> (3±1)/2 =>4/2 ,. 2/2. => 2 ,. 1. Hence: D= (-infinity , 1] , [ 2 , infinity).