Find Where Undefined/Discontinuous f(x)=(5x)/(x^3+5x^2+6x)

$f (x) = \frac{5 x}{x^{3} + 5 x^{2} + 6 x}$

Step 1

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

$x^{3} + 5 x^{2} + 6 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 5 x^{2} + 6 x$ .

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

$x \cdot x^{2} + 5 x^{2} + 6 x = 0$

Factor $x$ out of $5 x^{2}$ .

$x \cdot x^{2} + x (5 x) + 6 x = 0$

Factor $x$ out of $6 x$ .

$x \cdot x^{2} + x (5 x) + x \cdot 6 = 0$

Factor $x$ out of $x \cdot x^{2} + x (5 x)$ .

$x (x^{2} + 5 x) + x \cdot 6 = 0$

Factor $x$ out of $x (x^{2} + 5 x) + x \cdot 6$ .

$x (x^{2} + 5 x + 6) = 0$

Factor.

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

$x ((x + 2) (x + 3)) = 0$

Remove unnecessary parentheses.

$x (x + 2) (x + 3) = 0$

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

$x = 0$

$x + 2 = 0$

$x + 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 0, - 2, - 3$

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 $0$ .

$x = - 3, x = - 2, x = 0$

Step 4

Find Where Undefined/Discontinuous f(x)=(5x)/(x^3+5x^2+6x)

Related Questions

Support LCMGCF.com by disabling your adblocker.