Find Where Undefined/Discontinuous (4x)/(8x^2-4x)

$\frac{4 x}{8 x^{2} - 4 x}$

Step 1

Set the denominator in $\frac{4 x}{8 x^{2} - 4 x}$ equal to $0$ to find where the expression is undefined.

$8 x^{2} - 4 x = 0$

Step 2

Solve for $x$ .

Factor $4 x$ out of $8 x^{2} - 4 x$ .

Factor $4 x$ out of $8 x^{2}$ .

$4 x (2 x) - 4 x = 0$

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

$4 x (2 x) + 4 x (- 1) = 0$

Factor $4 x$ out of $4 x (2 x) + 4 x (- 1)$ .

$4 x (2 x - 1) = 0$

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

$x = 0$

$2 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

The final solution is all the values that make $4 x (2 x - 1) = 0$ true.

$x = 0, \frac{1}{2}$

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 $0$ .

$x = 0, x = \frac{1}{2}$

Step 4

Find Where Undefined/Discontinuous (4x)/(8x^2-4x)

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