Question

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

\\(x = -5\\)
\\(x = -2\\)
\\(x = 2\\)
\\(x = 5\\)

Explanation:

⚡ Using what you learned: Types of Discontinuity (removable, jump, infinite)

Step 1: Factor the denominator

Factor the quadratic expression in the denominator:

$$ x^2 + 3x - 10 = (x + 5)(x - 2) $$

Rewrite the function:

$$ f(x) = \frac{x + 5}{(x + 5)(x - 2)} $$

Step 2: Identify the removable discontinuity

A removable discontinuity occurs at a value of \( x \) that makes both the numerator and denominator zero, which allows the common factor to be canceled out.

The common factor is \( x + 5 \). Set this factor to zero:

$$ x + 5 = 0 $$

$$ x = -5 $$

Answer:

\( x = -5 \)