Find Where Undefined/Discontinuous (2y^3)/(2y^4+16y^3)

$\frac{2 y^{3}}{2 y^{4} + 16 y^{3}}$

Step 1

Set the denominator in $\frac{2 y^{3}}{2 y^{4} + 16 y^{3}}$ equal to $0$ to find where the expression is undefined.

$2 y^{4} + 16 y^{3} = 0$

Step 2

Solve for $y$ .

Factor $2 y^{3}$ out of $2 y^{4} + 16 y^{3}$ .

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

$2 y^{3} (y) + 16 y^{3} = 0$

Factor $2 y^{3}$ out of $16 y^{3}$ .

$2 y^{3} (y) + 2 y^{3} (8) = 0$

Factor $2 y^{3}$ out of $2 y^{3} (y) + 2 y^{3} (8)$ .

$2 y^{3} (y + 8) = 0$

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

$y^{3} = 0$

$y + 8 = 0$

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 $y + 8$ equal to $0$ and solve for $y$ .

Set $y + 8$ equal to $0$ .

$y + 8 = 0$

Subtract $8$ from both sides of the equation.

$y = - 8$

The final solution is all the values that make $2 y^{3} (y + 8) = 0$ true.

$y = 0, - 8$

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 $0$ .

$y = - 8, y = 0$

Step 4

Find Where Undefined/Discontinuous (2y^3)/(2y^4+16y^3)

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