where is the removable discontinuity of $f(x)=\frac{x + 5}{x^{2}+3x - 10}$ located?\n$x=-5$\n$x=-2$\n$x=2$\n$…

where is the removable discontinuity of $f(x)=\frac{x + 5}{x^{2}+3x - 10}$ located?\n$x=-5$\n$x=-2$\n$x=2$\n$x=5$

where is the removable discontinuity of $f(x)=\frac{x + 5}{x^{2}+3x - 10}$ located?\n$x=-5$\n$x=-2$\n$x=2$\n$x=5$

Answer

Explanation:

Step1: Factor the denominator

Factor $x^{2}+3x - 10$. We know that $x^{2}+3x - 10=(x + 5)(x-2)$ by using the formula $x^{2}+(a + b)x+ab=(x + a)(x + b)$ where $a = 5$ and $b=-2$. So $f(x)=\frac{x + 5}{(x + 5)(x - 2)}$.

Step2: Identify removable discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x + 5)$ can be canceled (for $x\neq - 5$). The function is undefined at the value of $x$ that makes the original denominator zero and is a removable - discontinuity when the common factor exists. Setting the common - factor $x+5 = 0$, we get $x=-5$.

Answer:

$x=-5$