find the value of a, if it exists, such that the following function is continuous at X = 2.
piecewise function
f(x)={x^2-3x+1, if x>2; a, if x=2; 3-2x, if x<2
I found the limits from left and right of 2 = -1 so the limit exists but I don't know how to figure out a
Am I missing something? If the limit from the left is -1, and the limit from the right is -1, and the function is continous, then the limit at the point is -1
So a=-1 That makes sense because it's continuous
to make it continuous f(2)=a so it would equal -1?
To determine the value of 'a' that makes the function continuous at x=2, we need to ensure that the function's left-hand limit, right-hand limit, and the value of the function at x=2 are all equal.
First, let's evaluate the left-hand limit (LHL) as x approaches 2:
lim (x->2-) (3-2x)
= 3 - 2(2)
= 3 - 4
= -1
Next, let's evaluate the right-hand limit (RHL) as x approaches 2:
lim (x->2+) (x^2 - 3x + 1)
= 2^2 - 3(2) + 1
= 4 - 6 + 1
= -1
From the above calculations, we can see that both the LHL and RHL are equal to -1.
Now, to ensure continuity, we equate the value of the function at x=2 to the limits we found:
a = -1
Therefore, the value of 'a' that makes the function continuous at x=2 is -1.