Find Where Undefined/Discontinuous (x^4+8x^2+7)/(3x^5-3x)

$\frac{x^{4} + 8 x^{2} + 7}{3 x^{5} - 3 x}$

Step 1

Set the denominator in $\frac{x^{4} + 8 x^{2} + 7}{3 x^{5} - 3 x}$ equal to $0$ to find where the expression is undefined.

$3 x^{5} - 3 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $3 x$ out of $3 x^{5} - 3 x$ .

Factor $3 x$ out of $3 x^{5}$ .

$3 x (x^{4}) - 3 x = 0$

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

$3 x (x^{4}) + 3 x (- 1) = 0$

Factor $3 x$ out of $3 x (x^{4}) + 3 x (- 1)$ .

$3 x (x^{4} - 1) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$3 x ({(x^{2})}^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$3 x ({(x^{2})}^{2} - 1^{2}) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 1$ .

$3 x ((x^{2} + 1) (x^{2} - 1)) = 0$

Factor.

Simplify.

Rewrite $1$ as $1^{2}$ .

$3 x ((x^{2} + 1) (x^{2} - 1^{2})) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$3 x ((x^{2} + 1) ((x + 1) (x - 1))) = 0$

Remove unnecessary parentheses.

$3 x ((x^{2} + 1) (x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$3 x (x^{2} + 1) (x + 1) (x - 1) = 0$

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

$x = 0$

$x^{2} + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

$x = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $3 x (x^{2} + 1) (x + 1) (x - 1) = 0$ true.

$x = 0, i, - i, - 1, 1$

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 $0$ .

$x = - 1, x = 0, x = 1$

Step 4

Find Where Undefined/Discontinuous (x^4+8x^2+7)/(3x^5-3x)

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