Find Where Undefined/Discontinuous f(x)=(2x^2-6x)/(x^2-4x+3)

$f (x) = \frac{2 x^{2} - 6 x}{x^{2} - 4 x + 3}$

Step 1

Set the denominator in $\frac{2 x^{2} - 6 x}{x^{2} - 4 x + 3}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 4 x + 3 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

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

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

$x - 3 = 0$

$x - 1 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

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 $0$ .

$x = 1, x = 3$

Step 4

Find Where Undefined/Discontinuous f(x)=(2x^2-6x)/(x^2-4x+3)

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