The graph below shows the function f(x)=5x+10/x^2+7x+10 Where is the removable discontinuity of f(x) located?
Accepted Solution
A:
For this case we have the following function: [tex] f (x) = (5x + 10) / (x ^ 2 + 7x + 10)
[/tex] Let's rewrite the function: [tex] f (x) = (5x + 10) / ((x + 2) (x + 5))
f (x) = (5 (x + 2)) / ((x + 2) (x + 5))[/tex] We set the denominator to zero to see the values of x for which it is not defined: [tex] (x + 2) (x + 5) = 0
[/tex] From here, we get: [tex]x = -2
x = -5
[/tex] There is a removable discontinuity at x = -2, since by rewriting the function we have: [tex] f (x) = 5 / (x + 5)
[/tex] Answer: the removable discontinuity of f (x) is located at: x = -2