Question

question
consider the graph of the function $f(x)=\frac{x^{2}+7x}{x^{3}-5x^{2}-14x}$.
where is the removable discontinuity? give only the $x$-value.
sorry, thats incorrect. try again?
i

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+7x=x(x + 7)$. The denominator $x^{3}-5x^{2}-14x=x(x^{2}-5x - 14)=x(x - 7)(x+2)$. So $f(x)=\frac{x(x + 7)}{x(x - 7)(x + 2)}$.

Step2: Identify the removable - discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $x$ cancels out (but $x
eq0$). When $x = 0$, the function has a removable discontinuity because the common factor $x$ can be removed after simplifying the function.

Answer:

$0$