Find Where Undefined/Discontinuous (x-7)/(2x^2-32)

$\frac{x - 7}{2 x^{2} - 32}$

Step 1

Set the denominator in $\frac{x - 7}{2 x^{2} - 32}$ equal to $0$ to find where the expression is undefined.

$2 x^{2} - 32 = 0$

Step 2

Solve for $x$ .

Add $32$ to both sides of the equation.

$2 x^{2} = 32$

Divide each term in $2 x^{2} = 32$ by $2$ and simplify.

Divide each term in $2 x^{2} = 32$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{32}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{32}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{32}{2}$

Simplify the right side.

Divide $32$ by $2$ .

$x^{2} = 16$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$

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 $0$ .

$x = - 4, x = 4$

Step 4

Find Where Undefined/Discontinuous (x-7)/(2x^2-32)

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