Find Where Undefined/Discontinuous f(x)=(x^3+3x^2-10x)/(x^3-25x)

$f (x) = \frac{x^{3} + 3 x^{2} - 10 x}{x^{3} - 25 x}$

Set the denominator in $\frac{x^{3} + 3 x^{2} - 10 x}{x^{3} - 25 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} - 25 x = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 25 x$ .

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

$x \cdot x^{2} - 25 x = 0$

Factor $x$ out of $- 25 x$ .

$x \cdot x^{2} + x \cdot - 25 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 25$ .

$x (x^{2} - 25) = 0$

Rewrite $25$ as $5^{2}$ .

$x (x^{2} - 5^{2}) = 0$

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 $b = 5$ .

$x ((x + 5) (x - 5)) = 0$

Remove unnecessary parentheses.

$x (x + 5) (x - 5) = 0$

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

$x = 0$

$x + 5 = 0$

$x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $x (x + 5) (x - 5) = 0$ true.

$x = 0, - 5, 5$

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 $0$ .

$x = - 5, x = 0, x = 5$

Find Where Undefined/Discontinuous f(x)=(x^3+3x^2-10x)/(x^3-25x)

Related Questions

Support LCMGCF.com by disabling your adblocker.