Find Where Undefined/Discontinuous f(x)=(x^2-5x-36)/(6x^2-13x+6)

$f (x) = \frac{x^{2} - 5 x - 36}{6 x^{2} - 13 x + 6}$

Set the denominator in $\frac{x^{2} - 5 x - 36}{6 x^{2} - 13 x + 6}$ equal to $0$ to find where the expression is undefined.

$6 x^{2} - 13 x + 6 = 0$

Solve for $x$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot 6 = 36$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$6 x^{2} - 13 x + 6 = 0$

Rewrite $- 13$ as $- 4$ plus $- 9$

$6 x^{2} + (- 4 - 9) x + 6 = 0$

Apply the distributive property.

$6 x^{2} - 4 x - 9 x + 6 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} - 4 x) - 9 x + 6 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

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

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

$3 x - 2 = 0$

$2 x - 3 = 0$

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

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

$x = \frac{2}{3}, \frac{3}{2}$

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 $0$ .

$x = \frac{2}{3}, x = \frac{3}{2}$

Find Where Undefined/Discontinuous f(x)=(x^2-5x-36)/(6x^2-13x+6)

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