Find Interval Where Function Is Continuous

[nycmathdad]

How do I find an interval where a given function is continuous? How is this done algebracially? How is this done using a graph?

Three sample function are given below.

1. f(x) = (x^2 - 9)/(x + 3)...Rational

2. g(x) = | x | + 2...Absolute value

3. h(x) = sqrt{16 - x^2}...square root

 

[HallsofIvy]

When x= -3, f(-3)= (9- 9)/(-3+ 3)= 0/0 so is not defined. But as long as x is not -3, (x^2- 9)/(x+ 3)= (x- 3)(x+ 3)/(x+ 3)= x- 3 which is always continuous. f is not continuous at x= -3 but is continuous on \(\displaystyle (-\infty, -3)\) and \(\displaystyle (-3, \infty)\).

Absolute value is continuous for all x so |x|+ 2 is continuous for all x: \(\displaystyle (-\infty, \infty)\).

16- x^2 is positive as long as \(\displaystyle 16- x^2> 0\), \(\displaystyle 16> x^2\), \(\displaystyle -4< x< 4\), so is sqrt exists and is continuous, but is negative, so the sqrt is not defined, for x< -4 or x> 4. Whether it is continuous at x= -4 or x= 4 depends on exactly how you define "continuous". You might want to consider "continuous from the left" and "continuous from the right".

 

[nycmathdad]

When x= -3, f(-3)= (9- 9)/(-3+ 3)= 0/0 so is not defined. But as long as x is not -3, (x^2- 9)/(x+ 3)= (x- 3)(x+ 3)/(x+ 3)= x- 3 which is always continuous. f is not continuous at x= -3 but is continuous on \(\displaystyle (-\infty, -3)\) and \(\displaystyle (-3, \infty)\).

Absolute value is continuous for all x so |x|+ 2 is continuous for all x: \(\displaystyle (-\infty, \infty)\).

16- x^2 is positive as long as \(\displaystyle 16- x^2> 0\), \(\displaystyle 16> x^2\), \(\displaystyle -4< x< 4\), so is sqrt exists and is continuous, but is negative, so the sqrt is not defined, for x< -4 or x> 4. Whether it is continuous at x= -4 or x= 4 depends on exactly how you define "continuous". You might want to consider "continuous from the left" and "continuous from the right".

Ok. I will practice by answering a few more questions.