Give the domain of the following functions. Give your answer in interval notation. a) f(x)= 5x-2/x^2-5x+6 and b) square root of 5-2x
a. F(x) = (5x-2)/(x^2-5x+6)
F(x) = (5x-2)/(x-2)(x-3).
Any value of x that causes a zero
denominator is outside of the domain.
X = 2 and X = 3 will cause the denominator to be o. Therefore, they are outside of the domain.
Domain: All real values of X except 2, and 3.
b. Sqrt(5-2x).
Any value of X that causes the radical to go negative is outside the domain.
2x > 5.
X > 5/2
X > 2.5 is outside of domain.
Domain: X <= 2.5. Or 2.5 > X =2.5.
To determine the domain of a function, we need to find the values of x for which the function is defined. In other words, we are looking for any restrictions or limitations on the values of x that can be plugged into the function.
a) For the function f(x) = (5x-2)/(x^2-5x+6), the only restriction to consider is the denominator (x^2-5x+6). To find the values of x for which the denominator is defined, we need to solve the equation x^2-5x+6 = 0.
Factoring the quadratic equation, we have (x-2)(x-3) = 0. Setting each factor equal to zero, we get x-2 = 0 and x-3 = 0, giving us x = 2 and x = 3.
Therefore, the function f(x) is not defined for x = 2 and x = 3, as these values would make the denominator equal to zero.
The domain of the function f(x) is all real numbers except x = 2 and x = 3, which can be expressed in interval notation as (-∞, 2) ∪ (2, 3) ∪ (3, ∞).
b) For the function f(x) = √(5-2x), we need to consider the values under the square root symbol. The expression inside the square root (5-2x) should be greater than or equal to zero to ensure that the function is defined.
Setting 5-2x ≥ 0 and solving for x, we get -2x ≥ -5, and upon dividing by -2 (remember to reverse the inequality sign since we are dividing by a negative number), we have x ≤ 5/2 or x ≤ 2.5.
Hence, the domain of the function f(x) is all real numbers less than or equal to 2.5, expressed in interval notation as (-∞, 2.5].
a) To find the domain of the function f(x) = (5x-2)/(x^2-5x+6), we need to consider the values of x for which the function is defined.
First, let's find the values of x that make the denominator zero, as dividing by zero is undefined. The denominator, x^2-5x+6, factors to (x-2)(x-3).
Setting each factor equal to zero, we have x-2=0 and x-3=0. Solving these equations, we find x=2 and x=3.
Therefore, the function is undefined for x=2 and x=3, so we exclude these values from the domain.
In interval notation, the domain of f(x) is (-∞, 2) ∪ (2, 3) ∪ (3, +∞).
b) The function f(x) = √(5-2x) represents the square root of (5-2x).
For the square root function, the radicand (expression inside the square root) must be greater than or equal to zero, as the square root of a negative number is undefined in real numbers.
Setting 5-2x ≥ 0, we can solve this inequality:
5-2x ≥ 0
-2x ≥ -5
Dividing both sides by -2 (remembering to flip the inequality when multiplying/dividing by a negative number), we get:
x ≤ 5/2
Therefore, the function is defined for x values less than or equal to 5/2.
In interval notation, the domain of f(x) is (-∞, 5/2].