What value must be defined for to remove the discontinuity of this function at ?
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Answer:
Option D. 8
Step-by-step explanation:
we have
[tex]f(x)=\frac{x^{2}-16 }{x-4}[/tex]
Remember that
[tex]x^{2}-16 =(x+4)(x-4)[/tex]
substitute
[tex]f(x)=\frac{(x+4)(x-4)}{x-4}[/tex]
Remove the discontinuity at x=4
Simplify
[tex]f(x)=(x+4)[/tex]
so
Find the value of f(4)
For x=4
[tex]f(4)=(4+4)=8[/tex]