Find the Domain and Range f(x) = log base 1/2 of |2-x|

$f (x) = \log_{\frac{1}{2}} (| 2 - x |)$

Step 1

Set the argument in $\log_{\frac{1}{2}} (| 2 - x |)$ greater than $0$ to find where the expression is defined.

$| 2 - x | > 0$

Step 2

Solve for $x$ .

Write $| 2 - x | > 0$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$2 - x \geq 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$- x \geq - 2$

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x \leq 2$

In the piece where $2 - x$ is non-negative, remove the absolute value.

$2 - x > 0$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$2 - x < 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$- x < - 2$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x > 2$

In the piece where $2 - x$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 - x) > 0$

Write as a piecewise.

${\begin{cases} 2 - x > 0 & x \leq 2 \\ - (2 - x) > 0 & x > 2 \end{cases}$

Simplify $- (2 - x) > 0$ .

Apply the distributive property.

${\begin{cases} 2 - x > 0 & x \leq 2 \\ - 1 \cdot 2 - - x > 0 & x > 2 \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} 2 - x > 0 & x \leq 2 \\ - 2 - - x > 0 & x > 2 \end{cases}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

${\begin{cases} 2 - x > 0 & x \leq 2 \\ - 2 + 1 x > 0 & x > 2 \end{cases}$

Multiply $x$ by $1$ .

${\begin{cases} 2 - x > 0 & x \leq 2 \\ - 2 + x > 0 & x > 2 \end{cases}$

Solve $2 - x > 0$ for $x$ .

Subtract $2$ from both sides of the inequality.

$- x > - 2$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x < 2$

Add $2$ to both sides of the inequality.

$x > 2$

Find the union of the solutions.

$x < 2$ or $x > 2$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 2}$

Step 4

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 2) \cup (2, \infty), {x | x \neq 2}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 6

Find the Domain and Range f(x) = log base 1/2 of |2-x|

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