Precalculus Examples

$f (x) = \frac{3 x^{2} + 2 x - 5}{2 x^{2} - 7 x}$

Step 1

Set the denominator in $\frac{3 x^{2} + 2 x - 5}{2 x^{2} - 7 x}$ equal to $0$ to find where the expression is undefined.

$2 x^{2} - 7 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x$ out of $2 x^{2} - 7 x$ .

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Step 2.1.1

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

$x (2 x) - 7 x = 0$

Step 2.1.2

Factor $x$ out of $- 7 x$ .

$x (2 x) + x \cdot - 7 = 0$

Step 2.1.3

Factor $x$ out of $x (2 x) + x \cdot - 7$ .

$x (2 x - 7) = 0$

Step 2.2

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

$x = 0$

$2 x - 7 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $2 x - 7$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Step 2.4.2

Solve $2 x - 7 = 0$ for $x$ .

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Step 2.4.2.1

Add $7$ to both sides of the equation.

$2 x = 7$

Step 2.4.2.2

Divide each term in $2 x = 7$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Step 2.5

The final solution is all the values that make $x (2 x - 7) = 0$ true.

$x = 0, \frac{7}{2}$

Step 3

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

Interval Notation:

$(- \infty, 0) \cup (0, \frac{7}{2}) \cup (\frac{7}{2}, \infty)$

Set-Builder Notation:

${x | x \neq 0, \frac{7}{2}}$

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: $(- \infty, 0) \cup (0, \frac{7}{2}) \cup (\frac{7}{2}, \infty), {x | x \neq 0, \frac{7}{2}}$

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

Step 6