Find Where Undefined/Discontinuous y=1/((x-2)(x+3))

$y = \frac{1}{(x - 2) (x + 3)}$

Step 1

Set the denominator in $\frac{1}{(x - 2) (x + 3)}$ equal to $0$ to find where the expression is undefined.

$(x - 2) (x + 3) = 0$

Step 2

Solve for $x$ .

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

$x - 2 = 0$

$x + 3 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

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 $0$ .

$x = - 3, x = 2$

Step 4

Find Where Undefined/Discontinuous y=1/((x-2)(x+3))

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