where is the removable discontinuity of f(x) = (x + 5)/(x^2+3x - 10) located?\no x = -5\no x = -2\no x =…

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 $ax^{2}+bx + c=a(x - x_1)(x - x_2)$ where $x_1,x_2$ are the roots of the quadratic equation $ax^{2}+bx + c = 0$. For $x^{2}+3x - 10 = 0$, $a = 1$, $b=3$, $c=-10$, and the roots are $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-3\pm\sqrt{9+40}}{2}=\frac{-3\pm7}{2}$, which gives $x=-5$ and $x = 2$. So $f(x)=\frac{x + 5}{(x + 5)(x - 2)}$.

Step2: Simplify the function

Cancel out the common - factor $(x + 5)$ (for $x\neq - 5$), and we get $f(x)=\frac{1}{x - 2},x\neq - 5$. A removable discontinuity occurs when a factor in the numerator and denominator cancels out. The factor $x + 5$ cancels, and the value of $x$ that makes this factor zero is $x=-5$.

Answer:

$x=-5$