Find Where Undefined/Discontinuous (-4x)/(8x^2-3x)

$\frac{- 4 x}{8 x^{2} - 3 x}$

Step 1

Set the denominator in $\frac{- 4 x}{8 x^{2} - 3 x}$ equal to $0$ to find where the expression is undefined.

$8 x^{2} - 3 x = 0$

Step 2

Solve for $x$ .

Factor $x$ out of $8 x^{2} - 3 x$ .

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

$x (8 x) - 3 x = 0$

Factor $x$ out of $- 3 x$ .

$x (8 x) + x \cdot - 3 = 0$

Factor $x$ out of $x (8 x) + x \cdot - 3$ .

$x (8 x - 3) = 0$

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

$x = 0$

$8 x - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $8 x - 3$ equal to $0$ and solve for $x$ .

Set $8 x - 3$ equal to $0$ .

$8 x - 3 = 0$

Solve $8 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$8 x = 3$

Divide each term in $8 x = 3$ by $8$ and simplify.

Divide each term in $8 x = 3$ by $8$ .

$\frac{8 x}{8} = \frac{3}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{3}{8}$

Divide $x$ by $1$ .

$x = \frac{3}{8}$

The final solution is all the values that make $x (8 x - 3) = 0$ true.

$x = 0, \frac{3}{8}$

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 $0$ .

$x = 0, x = \frac{3}{8}$

Step 4

Find Where Undefined/Discontinuous (-4x)/(8x^2-3x)

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