Find the Domain of the Product of the Functions f(x)=1/x , g(x) = square root of 2-5x

$f (x) = \frac{1}{x}$ , $g (x) = \sqrt{2 - 5 x}$

Step 1

Find the product of the functions.

Replace the function designators with the actual functions in $f (x) \cdot (g (x))$ .

$(\frac{1}{x}) \cdot (\sqrt{2 - 5 x})$

Combine $\frac{1}{x}$ and $\sqrt{2 - 5 x}$ .

$\frac{\sqrt{2 - 5 x}}{x}$

Step 2

Set the radicand in $\sqrt{2 - 5 x}$ greater than or equal to $0$ to find where the expression is defined.

$2 - 5 x \geq 0$

Step 3

Solve for $x$ .

Subtract $2$ from both sides of the inequality.

$- 5 x \geq - 2$

Divide each term in $- 5 x \geq - 2$ by $- 5$ and simplify.

Divide each term in $- 5 x \geq - 2$ by $- 5$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 5 x}{- 5} \leq \frac{- 2}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} \leq \frac{- 2}{- 5}$

Divide $x$ by $1$ .

$x \leq \frac{- 2}{- 5}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{2}{5}$

Step 4

Set the denominator in $\frac{\sqrt{2 - 5 x}}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 5

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

Interval Notation:

$(- \infty, 0) \cup (0, \frac{2}{5}]$

Set-Builder Notation:

${x | x \leq \frac{2}{5}, x \neq 0}$

Step 6

Find the Domain of the Product of the Functions f(x)=1/x , g(x) = square root of 2-5x

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