Find the Domain and Range f(x) = square root of x+ square root of 1-x

$f (x) = \sqrt{x} + \sqrt{1 - x}$

Step 1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2

Set the radicand in $\sqrt{1 - x}$ greater than or equal to $0$ to find where the expression is defined.

$1 - x \geq 0$

Step 3

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

Divide each term in $- x \geq - 1$ by $- 1$ and simplify.

Divide each term in $- x \geq - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x \leq 1$

Step 4

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, 1]$

Set-Builder Notation:

${x | 0 \leq x \leq 1}$

Step 5

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[1, \sqrt{2}]$

Set-Builder Notation:

${y | 1 \leq y \leq \sqrt{2}}$

Step 6

Determine the domain and range.

Domain: $[0, 1], {x | 0 \leq x \leq 1}$

Range: $[1, \sqrt{2}], {y | 1 \leq y \leq \sqrt{2}}$

Step 7

Find the Domain and Range f(x) = square root of x+ square root of 1-x

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