where is the removable discontinuity of f(x)=(x + 5)/(x^2+3x - 10) located?\no x=-5\no x=-2\no x=2\no x=5
where is the removable discontinuity of f(x)=(x + 5)/(x^2+3x - 10) located?\no x=-5\no x=-2\no x=2\no x=5
Answer
Explanation:
Step1: Factor the denominator
Factor $x^{2}+3x - 10=(x + 5)(x - 2)$. 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 $(x + 5)$ cancels out (for $x\neq - 5$). The function is undefined at the values that make the denominator zero. After canceling the common factor $(x + 5)$, the original discontinuity at $x=-5$ is removable.
Answer:
A. $x = - 5$