Find Where Undefined/Discontinuous (-6x)/(9x^2-4x)

$\frac{- 6 x}{9 x^{2} - 4 x}$

Step 1

Set the denominator in $\frac{- 6 x}{9 x^{2} - 4 x}$ equal to $0$ to find where the expression is undefined.

$9 x^{2} - 4 x = 0$

Step 2

Solve for $x$ .

Factor $x$ out of $9 x^{2} - 4 x$ .

Factor $x$ out of $9 x^{2}$ .

$x (9 x) - 4 x = 0$

Factor $x$ out of $- 4 x$ .

$x (9 x) + x \cdot - 4 = 0$

Factor $x$ out of $x (9 x) + x \cdot - 4$ .

$x (9 x - 4) = 0$

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

$x = 0$

$9 x - 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $9 x - 4$ equal to $0$ and solve for $x$ .

Set $9 x - 4$ equal to $0$ .

$9 x - 4 = 0$

Solve $9 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$9 x = 4$

Divide each term in $9 x = 4$ by $9$ and simplify.

Divide each term in $9 x = 4$ by $9$ .

$\frac{9 x}{9} = \frac{4}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x}{9} = \frac{4}{9}$

Divide $x$ by $1$ .

$x = \frac{4}{9}$

The final solution is all the values that make $x (9 x - 4) = 0$ true.

$x = 0, \frac{4}{9}$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = 0, x = \frac{4}{9}$

Step 4

Find Where Undefined/Discontinuous (-6x)/(9x^2-4x)

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