F(x)=4 if x=2
how do i show that f(x) is discontinuous at x=2 and how do i remove the discontinuity by changing f(2)?
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Correction: f(x)=sqrt 4x+1 if x is not equal to 2
Square root functions h(x) = sqrt (x) are continuous. And linear functions like g(x) = 4x + 1 are also continuous. So you'd expect that when you compose them with each other to get f(x) = sqrt(4x +1), you would have f(x) be continuous whenever the function is defined. This is actually a theorem in math.
So if you imagine defining f(x) as sqrt(4x + 1) for all x not equal to 2, and then f(x) = 4 when x = 4, to check whether f(x) is continuous at 2, all you would want to check is to see whether or not sqrt(4*2 + 1) = f(2) = 4.
If sqrt(4*2 + 1) = 4. then f(x) is continuous. If they're not equal, then f(x) is discontinuous at 2.
How would you remove it? Well sqrt(4x + 1) is a continuous function,
so you'd just want f(2) to be whatever sqrt(4x + 1) is when x = 2.
So change f(2) to sqrt(4*2 + 1).
The theorem that compositions of continuous functions are continuous might not have been discussed in your class yet. In that case, you'd first want to check the left and right hand limits of f(x) as x approaches 2.
If both left and right hand limits are equal, then the function approaches that limit (which turns out to be 3). To see whether the function is continuous, see whether the function is equal to the limit at that point (4 isn't equal to 3). Then you'd remove the discontinuity like above.
So if you imagine defining f(x) as sqrt(4x + 1) for all x not equal to 2, and then f(x) = 4 when x = *2* ...