Find the Domain and Range f(x)=(2+x)/(x^2-7x)

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

Step 1

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

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

Step 2

Solve for $x$ .

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

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

$x \cdot x - 7 x = 0$

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

$x \cdot x + x \cdot - 7 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$x (x - 7) = 0$

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

$x = 0$

$x - 7 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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

$x = 0, 7$

Step 3

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

Interval Notation:

$(- \infty, 0) \cup (0, 7) \cup (7, \infty)$

Set-Builder Notation:

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

Step 4

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

Interval Notation:

$(- \infty, - \frac{11 + 6 \sqrt{2}}{49}] \cup [- \frac{11 - 6 \sqrt{2}}{49}, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{11 + 6 \sqrt{2}}{49}, y \geq - \frac{11 - 6 \sqrt{2}}{49}}$

Step 5

Determine the domain and range.

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

Range: $(- \infty, - \frac{11 + 6 \sqrt{2}}{49}] \cup [- \frac{11 - 6 \sqrt{2}}{49}, \infty), {y | y \leq - \frac{11 + 6 \sqrt{2}}{49}, y \geq - \frac{11 - 6 \sqrt{2}}{49}}$

Step 6

Find the Domain and Range f(x)=(2+x)/(x^2-7x)

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