find domain of (x-5)^(1/3)/(x^3-5)
To find the domain of the given expression, we need to identify any values of x that would result in division by zero or taking an even root of a negative number.
First, let's consider the denominator, x^3 - 5. The expression would be undefined if x^3 - 5 = 0. Solving for x:
x^3 = 5
x = ∛5
Next, let's consider the numerator, (x - 5)^(1/3). Taking the cube root of a negative number would result in a complex number, so we need to make sure x - 5 is not negative.
x - 5 ≥ 0
x ≥ 5
Therefore, the domain of the expression is all real numbers greater than or equal to 5, or [5, ∞).
To find the domain of the given expression, we need to determine the values of x for which the expression is defined.
First, let's consider the denominator, x^3 - 5. This expression is defined for all real values of x, except when the denominator becomes zero. So, we need to find the values of x that make the denominator equal to zero.
Setting the denominator to zero, we have:
x^3 - 5 = 0
To solve for x, we can add 5 to both sides of the equation:
x^3 = 5
Taking the cube root of both sides:
x = ∛5
Therefore, x = ∛5 is a value for which the denominator becomes zero.
Now, let's consider the numerator, (x - 5)^(1/3). This expression is defined for all real values of x.
To determine the overall domain, we need to exclude the value of x = ∛5 (for which the denominator becomes zero) from the set of real numbers.
Therefore, the domain of the given expression is all real numbers except x = ∛5.
To find the domain of the expression, we need to consider any values of x that would make the expression undefined. In this case, we need to consider two things:
1. The denominator cannot be equal to zero.
2. The index of the radical (in this case, the cube root) should be valid.
First, let's consider the denominator: x^3 - 5 ≠ 0
To solve this equation, we can set the denominator equal to zero and solve for x:
x^3 - 5 = 0
By adding 5 to both sides of the equation, we get:
x^3 = 5
Taking the cube root of both sides, we find:
x = ∛5
So, one value that would make the denominator zero is x = ∛5.
Next, let's consider the index of the radical, which is 3 in this case. The index should be valid, meaning the expression inside the radical should be non-negative:
x - 5 ≥ 0
Solving this inequality, we find:
x ≥ 5
Combining this result with the previous value of x = ∛5, we find that the expression is undefined when x < ∛5 or x = ∛5.
Therefore, the domain of the expression is all real numbers greater than or equal to ∛5. In interval notation, the domain can be written as [∛5, +∞).