Question

1. find the x-values where the function is not continuous and classify the discontinuity as removable or non-removable. given: ( y = \frac{x + 5}{x^2 + 2x - 15} )

Explanation:

Step1: Factor the denominator

To find where the function is discontinuous, we first factor the denominator \(x^2 + 2x - 15\). We need two numbers that multiply to \(-15\) and add to \(2\). Those numbers are \(5\) and \(-3\). So, \(x^2 + 2x - 15=(x + 5)(x - 3)\). The function becomes \(y=\frac{x + 5}{(x + 5)(x - 3)}\).

Step2: Find the values that make the denominator zero

A function is discontinuous where the denominator is zero (since division by zero is undefined). So we set the denominator equal to zero: \((x + 5)(x - 3)=0\). Using the zero - product property, we have \(x+5 = 0\) or \(x - 3=0\). Solving these equations gives \(x=-5\) or \(x = 3\).

Step3: Simplify the function and classify discontinuities

We can simplify the function \(y=\frac{x + 5}{(x + 5)(x - 3)}\) (for \(x
eq - 5\)) by canceling out the common factor \((x + 5)\) in the numerator and the denominator. The simplified function is \(y=\frac{1}{x - 3}\) (for \(x
eq - 5\)).

• For \(x=-5\): When we cancel the \((x + 5)\) terms, we are essentially "removing" the discontinuity at \(x=-5\) because the limit as \(x\) approaches \(-5\) of the original function exists. We can find the limit: \(\lim_{x

ightarrow - 5}\frac{x + 5}{(x + 5)(x - 3)}=\lim_{x
ightarrow - 5}\frac{1}{x - 3}=\frac{1}{-5 - 3}=-\frac{1}{8}\). So the discontinuity at \(x = - 5\) is removable.

• For \(x = 3\): The denominator of the simplified function \(y=\frac{1}{x - 3}\) is zero when \(x = 3\), and the limit as \(x\) approaches \(3\) of \(\frac{1}{x - 3}\) does not exist (it approaches \(\pm\infty\)). So the discontinuity at \(x=3\) is non - removable.

Answer:

The function is not continuous at \(x=-5\) (removable discontinuity) and \(x = 3\) (non - removable discontinuity).