Find Where Undefined/Discontinuous (y^4+5y^2+6)/(4y^5+12y^3)

$\frac{y^{4} + 5 y^{2} + 6}{4 y^{5} + 12 y^{3}}$

Step 1

Set the denominator in $\frac{y^{4} + 5 y^{2} + 6}{4 y^{5} + 12 y^{3}}$ equal to $0$ to find where the expression is undefined.

$4 y^{5} + 12 y^{3} = 0$

Step 2

Solve for $y$ .

Factor $4 y^{3}$ out of $4 y^{5} + 12 y^{3}$ .

Factor $4 y^{3}$ out of $4 y^{5}$ .

$4 y^{3} (y^{2}) + 12 y^{3} = 0$

Factor $4 y^{3}$ out of $12 y^{3}$ .

$4 y^{3} (y^{2}) + 4 y^{3} (3) = 0$

Factor $4 y^{3}$ out of $4 y^{3} (y^{2}) + 4 y^{3} (3)$ .

$4 y^{3} (y^{2} + 3) = 0$

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

$y^{3} = 0$

$y^{2} + 3 = 0$

Set $y^{3}$ equal to $0$ and solve for $y$ .

Set $y^{3}$ equal to $0$ .

$y^{3} = 0$

Solve $y^{3} = 0$ for $y$ .

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

$y = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$y = \sqrt[3]{0^{3}}$

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

$y = 0$

Set $y^{2} + 3$ equal to $0$ and solve for $y$ .

Set $y^{2} + 3$ equal to $0$ .

$y^{2} + 3 = 0$

Solve $y^{2} + 3 = 0$ for $y$ .

Subtract $3$ from both sides of the equation.

$y^{2} = - 3$

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

$y = \pm \sqrt{- 3}$

Simplify $\pm \sqrt{- 3}$ .

Rewrite $- 3$ as $- 1 (3)$ .

$y = \pm \sqrt{- 1 (3)}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$y = \pm \sqrt{- 1} \cdot \sqrt{3}$

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

$y = \pm i \sqrt{3}$

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.

$y = i \sqrt{3}$

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

$y = - i \sqrt{3}$

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

$y = i \sqrt{3}, - i \sqrt{3}$

The final solution is all the values that make $4 y^{3} (y^{2} + 3) = 0$ true.

$y = 0, i \sqrt{3}, - i \sqrt{3}$

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 $0$ .

$y = 0$

Step 4

Find Where Undefined/Discontinuous (y^4+5y^2+6)/(4y^5+12y^3)

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