Find the Domain and Range f(x) = square root of 2+x-x^2

$f (x) = \sqrt{2 + x - x^{2}}$

Step 1

Set the radicand in $\sqrt{2 + x - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$2 + x - x^{2} \geq 0$

Step 2

Solve for $x$ .

Convert the inequality to an equation.

$2 + x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $2 + x - x^{2}$ .

Reorder the expression.

Move $2$ .

$x - x^{2} + 2 = 0$

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 2 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 2 = 0$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 2 = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 2)$ .

$- (x^{2} - x - 2) = 0$

Factor.

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$- ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

$x + 1 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$x > 2$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$2 - 4 - {(- 4)}^{2} \geq 0$

The left side $- 18$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$2 + 0 - {(0)}^{2} \geq 0$

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$2 + 4 - {(4)}^{2} \geq 0$

The left side $- 10$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

$x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 1 \leq x \leq 2$

Step 3

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

Interval Notation:

$[- 1, 2]$

Set-Builder Notation:

${x | - 1 \leq x \leq 2}$

Step 4

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

Interval Notation:

$[0, \frac{3}{2}]$

Set-Builder Notation:

${y | 0 \leq y \leq \frac{3}{2}}$

Step 5

Determine the domain and range.

Domain: $[- 1, 2], {x | - 1 \leq x \leq 2}$

Range: $[0, \frac{3}{2}], {y | 0 \leq y \leq \frac{3}{2}}$

Step 6

Find the Domain and Range f(x) = square root of 2+x-x^2

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