Find Where Undefined/Discontinuous (x^2-36)/(x^2)*(x^2-6x)/(x^2+x-42)

$\frac{x^{2} - 36}{x^{2}} \cdot \frac{x^{2} - 6 x}{x^{2} + x - 42}$

Step 1

Set the denominator in $\frac{x^{2} - 36}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 2

Solve for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 3

Set the denominator in $\frac{x^{2} - 6 x}{x^{2} + x - 42}$ equal to $0$ to find where the expression is undefined.

$x^{2} + x - 42 = 0$

Step 4

Solve for $x$ .

Factor $x^{2} + x - 42$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$(x - 6) (x + 7) = 0$

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

$x - 6 = 0$

$x + 7 = 0$

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 6) (x + 7) = 0$ true.

$x = 6, - 7$

Step 5

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 $0$ .

$x = - 7, x = 0, x = 6$

Step 6

Find Where Undefined/Discontinuous (x^2-36)/(x^2)*(x^2-6x)/(x^2+x-42)

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