Find Where Undefined/Discontinuous f(x)=(x+1)/(6x^2-7x-3)

$f (x) = \frac{x + 1}{6 x^{2} - 7 x - 3}$

Step 1

Set the denominator in $\frac{x + 1}{6 x^{2} - 7 x - 3}$ equal to $0$ to find where the expression is undefined.

$6 x^{2} - 7 x - 3 = 0$

Step 2

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 $a \cdot c = 6 \cdot - 3 = - 18$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$6 x^{2} - 7 x - 3 = 0$

Rewrite $- 7$ as $2$ plus $- 9$

$6 x^{2} + (2 - 9) x - 3 = 0$

Apply the distributive property.

$6 x^{2} + 2 x - 9 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} + 2 x) - 9 x - 3 = 0$

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

$2 x (3 x + 1) - 3 (3 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$(3 x + 1) (2 x - 3) = 0$

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

$3 x + 1 = 0$

$2 x - 3 = 0$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

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 $(3 x + 1) (2 x - 3) = 0$ true.

$x = - \frac{1}{3}, \frac{3}{2}$

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 $0$ .

$x = - \frac{1}{3}, x = \frac{3}{2}$

Step 4

Find Where Undefined/Discontinuous f(x)=(x+1)/(6x^2-7x-3)

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