Question

Find the domain of the function. $$ f(x)=\frac{\sqrt[3]{x^{2}-x+1}}{x^{2}+1} $$

Short answer

The domain of the function \(f(x)=\frac{\sqrt[3]{x^{2}-x+1}}{x^{2}+1}\) is all real numbers, denoted as \((-∞, ∞)\).

Step-by-step solution

Examine the numerator

The numerator consists of a cubic root function, which is defined for all real numbers. Therefore, there are no restrictions on x-values from the numerator. Step 2: Examine the denominator

Examine the denominator

We need to check if there are any x-values that make the denominator equal to zero. Given the denominator is \(x^2 + 1\), let's find the values of x for which this expression is equal to zero: $$x^2 + 1 = 0$$ $$x^2 = -1$$ We can see that there is no real value of x for which \(x^2 = -1\). Therefore, the denominator is never zero for any real value of x. Step 3: Determine the domain

Determine the domain

Since there are no restrictions on x-values from the numerator and the denominator is never zero for any real value of x, the domain of the function is all real numbers. In interval notation, the domain of the function \(f(x)=\frac{\sqrt[3]{x^{2}-x+1}}{x^{2}+1}\) is written as: $$(-\infty, \infty)$$