Precalculus Examples

$f (x) = \frac{\sqrt{2 - x}}{x^{2} - 4 x + 3}$

Step 1

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

$2 - x \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $2$ from both sides of the inequality.

$- x \geq - 2$

Step 2.2

Divide each term in $- x \geq - 2$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- x \geq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.2.2.2

Divide $x$ by $1$ .

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 2$ by $- 1$ .

$x \leq 2$

Step 3

Set the denominator in $\frac{\sqrt{2 - x}}{x^{2} - 4 x + 3}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 4 x + 3 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor $x^{2} - 4 x + 3$ using the AC method.

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Step 4.1.1

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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Step 4.1.2

Write the factored form using these integers.

$(x - 3) (x - 1) = 0$

Step 4.2

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

$x - 3 = 0$

$x - 1 = 0$

Step 4.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.4

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

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Step 4.4.1

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

$x - 1 = 0$

Step 4.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.5

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Step 5

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

Interval Notation:

$(- \infty, 1) \cup (1, 2]$

Set-Builder Notation:

${x | x \leq 2, x \neq 1}$

Step 6