Question

if f(x) ={x2-x-6/x2-2x-3}x=3,

find k so that the function f may be continuous at x = 3.​

Answer 1

Answer:

$\frac{ {x}^{2} - x - 6}{ {x}^{2} - 2x - 3 } \\ = \frac{ {x}^{2} - 3x + 2x - 6 }{ {x}^{2} - 3x + x - 3 } \\ = \frac{x(x - 3) + 2(x - 3)}{x(x - 3) + 1(x - 3)} \\ = \frac{( \cancel{ x - 3})(x + 2)}{( \cancel{ x - 3})(x + 1)} \\ = \frac{x + 2}{x + 1}$