Question

Find the domain of the function. $$ f(x)=\sqrt{x}+\sqrt{1-x} $$

Short answer

The domain of the function \(f(x)=\sqrt{x}+\sqrt{1-x}\) is \(x ∈ [0,1]\), i.e., from 0 to 1 inclusive.

Step-by-step solution

Identify the range of X for the first square root

The expression under the first square root is \(x\), and as we know the expression inside the square root should always be greater than or equal to zero. Thus \(x ≥ 0 \)

Identify the range of X for the second square root

The expression under the second square root is \(1-x\), hence the range of x for which it is greater than or equal to zero is \(1 - x ≥ 0 \). Simplifying it after making \(x\) subject, we get \(x ≤ 1 \)

Determine the common range of X for both square roots

Now that we've determined the range of x separately for both square roots, we need to find the common set of x-values that satisfies \(x ≥ 0 \) and \(x ≤ 1\). From both, it is clear that x is in the closed interval [0,1], i.e., \(0 ≤ x ≤ 1\)