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

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

Answer:

A. $x = - 5$

Explanation:

Step1: Factor the denominator

$x^{2}+3x - 10=(x + 5)(x - 2)$ 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 value of $x$ that makes this common - factor zero is $x=-5$. So the removable discontinuity is at $x = - 5$.