Precalculus Examples

$h (k) = k + \sqrt{6 - 4 k} - \sqrt[3]{k - 7}$

Step 1

Set the radicand in $\sqrt{6 - 4 k}$ greater than or equal to $0$ to find where the expression is defined.

$6 - 4 k \geq 0$

Step 2

Solve for $k$ .

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Step 2.1

Subtract $6$ from both sides of the inequality.

$- 4 k \geq - 6$

Step 2.2

Divide each term in $- 4 k \geq - 6$ by $- 4$ and simplify.

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Step 2.2.1

Divide each term in $- 4 k \geq - 6$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 k}{- 4} \leq \frac{- 6}{- 4}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 4$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 4 k}{- 4} \leq \frac{- 6}{- 4}$

Step 2.2.2.1.2

Divide $k$ by $1$ .

$k \leq \frac{- 6}{- 4}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Cancel the common factor of $- 6$ and $- 4$ .

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Step 2.2.3.1.1

Factor $- 2$ out of $- 6$ .

$k \leq \frac{- 2 (3)}{- 4}$

Step 2.2.3.1.2

Cancel the common factors.

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Step 2.2.3.1.2.1

Factor $- 2$ out of $- 4$ .

$k \leq \frac{- 2 \cdot 3}{- 2 \cdot 2}$

Step 2.2.3.1.2.2

Cancel the common factor.

$k \leq \frac{- 2 \cdot 3}{- 2 \cdot 2}$

Step 2.2.3.1.2.3

Rewrite the expression.

$k \leq \frac{3}{2}$

Step 3

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

Interval Notation:

$(- \infty, \frac{3}{2}]$

Set-Builder Notation:

${k | k \leq \frac{3}{2}}$

Step 4