Question

What value must be defined for to remove the discontinuity of this function at ?

Answer 1

Answer:

Option D. 8

Step-by-step explanation:

we have

$f(x)=\frac{x^{2}-16 }{x-4}$

Remember that

$x^{2}-16 =(x+4)(x-4)$

substitute

$f(x)=\frac{(x+4)(x-4)}{x-4}$

Remove the discontinuity at x=4

Simplify

$f(x)=(x+4)$

so

Find the value of f(4)

For x=4

$f(4)=(4+4)=8$