Find Where Undefined/Discontinuous (9x)/(54x^2-9x)

$\frac{9 x}{54 x^{2} - 9 x}$

Step 1

Set the denominator in $\frac{9 x}{54 x^{2} - 9 x}$ equal to $0$ to find where the expression is undefined.

$54 x^{2} - 9 x = 0$

Step 2

Solve for $x$ .

Factor $9 x$ out of $54 x^{2} - 9 x$ .

Factor $9 x$ out of $54 x^{2}$ .

$9 x (6 x) - 9 x = 0$

Factor $9 x$ out of $- 9 x$ .

$9 x (6 x) + 9 x (- 1) = 0$

Factor $9 x$ out of $9 x (6 x) + 9 x (- 1)$ .

$9 x (6 x - 1) = 0$

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

$x = 0$

$6 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $6 x - 1$ equal to $0$ and solve for $x$ .

Set $6 x - 1$ equal to $0$ .

$6 x - 1 = 0$

Solve $6 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$6 x = 1$

Divide each term in $6 x = 1$ by $6$ and simplify.

Divide each term in $6 x = 1$ by $6$ .

$\frac{6 x}{6} = \frac{1}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{1}{6}$

Divide $x$ by $1$ .

$x = \frac{1}{6}$

The final solution is all the values that make $9 x (6 x - 1) = 0$ true.

$x = 0, \frac{1}{6}$

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 $0$ .

$x = 0, x = \frac{1}{6}$

Step 4

Find Where Undefined/Discontinuous (9x)/(54x^2-9x)

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