How do you prove that the function #f(x)=(x^2+7x+10)/(x+5) # is continuous everywhere but x=-5?

Question

Charles Anctil · Accepted Answer

Factor and rewrite the function.

#f(x) = (x^2+7x+10)/(x+5)# # = ((x+2)(x+5))/(x+5)# # = x+2# for #x != -5# Use whatever tools you have developed to show that a linear function is continuous, but note that #f(-5)# is not defined, so #f# cannot be continuous at #-5#.

Charles Arocha · Answer

undefined To prove that the function f(x)=(x^2+7x+10)/(x+5) is continuous everywhere except at x=-5, we need to show that it satisfies the conditions for continuity.

First, we check if the function is defined at x=-5. Since dividing by zero is undefined, the function is not defined at x=-5.

Next, we examine the limit of the function as x approaches -5 from both sides. Taking the limit as x approaches -5 from the left side, we substitute x=-5 into the function and simplify:

lim(x→-5-) [(x^2+7x+10)/(x+5)] = (-5^2+7(-5)+10)/(-5+5) = (-25-35+10)/0 = -50/0

Similarly, taking the limit as x approaches -5 from the right side, we substitute x=-5 into the function and simplify:

lim(x→-5+) [(x^2+7x+10)/(x+5)] = (-5^2+7(-5)+10)/(-5+5) = (-25-35+10)/0 = -50/0

Since both limits result in -50/0, which is undefined, the limit of the function as x approaches -5 does not exist.

Therefore, we can conclude that the function f(x)=(x^2+7x+10)/(x+5) is continuous everywhere except at x=-5.