# Let f be given by the formula?

## Let f be given by the formula: Find the value(s) of x at which f is discontinuous?

Mar 1, 2018

At $x = 1$

#### Explanation:

Consider the denominator.
${x}^{2} + 2 x - 3$
Can be written as:
${x}^{2} + 2 x + 1 - 4$
${\left(x + 1\right)}^{2} - 4$
${\left(x + 1\right)}^{2} - {2}^{2}$

Now from relation ${a}^{2} - {b}^{2}$ = $\left(a + b\right) \left(a - b\right)$ we have
(x+1 +2)(x+1 -2))

(x+3)(x-1))

If $x = 1$, the denominator in above function is zero and the function tends to $\infty$ and not differentiable. Is discontinous.

Mar 1, 2018

$f \left(x\right) = \frac{x + 2}{{x}^{2} + 2 x - 3}$ is discontinuous when $x = - 3$ and $x = 1$

#### Explanation:

$f \left(x\right) = \frac{x + 2}{{x}^{2} + 2 x - 3}$ is discontinuous when denominator is zero i.e.

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

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

or $x \left(x + 3\right) - 1 \left(x + 3\right) = 0$

or $\left(x - 1\right) \left(x + 3\right) = 0$

i.e. $x = - 3$ and $x = 1$

graph{(x+2)/(x^2+2x-3) [-10, 10, -5, 5]}