Find the value of the constant C that makes the following function continuous on (-infinity, +inf)
f(x)={cx+2 if x is in (-inf,2]
cx^2-2 if x i in (2, inf)
cx+2 = cx^2-2 at x = 2
2c+2 = 4c-2
2c = 4
c = 2
Ah, the mysterious constant C strikes again! Let's figure out what value it needs to take in order to make this function continuous.
To make f(x) continuous, we need to ensure that the left-hand limit as x approaches 2 is equal to the right-hand limit as x approaches 2. So, let's calculate both of these limits separately.
First, let's find the left-hand limit:
lim (x -> 2-) f(x) = lim (x -> 2-) (cx+2)
Since we're approaching 2 from the left side, we can substitute x = 2 into the expression:
lim (x -> 2-) f(x) = lim (x -> 2-) (2c+2) = 2c + 2
Next, let's find the right-hand limit:
lim (x -> 2+) f(x) = lim (x -> 2+) (cx^2-2)
Again, we substitute x = 2 into the expression:
lim (x -> 2+) f(x) = lim (x -> 2+) (c * 2^2-2) = lim (x -> 2+) (4c-2) = 4c - 2
For f(x) to be continuous, the left-hand limit should be equal to the right-hand limit. Therefore, we have the equation:
2c + 2 = 4c - 2
Let's solve for c:
4c - 2 - 2c = 2
2c = 4
c = 2
So, the value of the constant C that makes the function continuous on (-∞, +∞) is 2.
To find the value of the constant C that makes the function continuous on the interval (-∞, +∞), we need to ensure that the two pieces of the function, cx + 2 and cx^2 - 2, connect smoothly at x = 2.
First, let's analyze the left-hand limit as x approaches 2 from the left side:
lim(x->2-) (cx + 2) = c * 2 + 2
Next, let's analyze the right-hand limit as x approaches 2 from the right side:
lim(x->2+) (cx^2 - 2) = c * 2^2 - 2
To make the function continuous at x = 2, these two limits should be equal. Therefore, we have:
c * 2 + 2 = c * 4 - 2
Simplifying the equation, we get:
2c + 2 = 4c - 2
Rearranging the equation, we have:
2 = 2c
Dividing both sides by 2, we find:
c = 1
Thus, the value of the constant C that makes the function continuous on the interval (-∞, +∞) is C = 1.
To find the value of the constant C that makes the function continuous on the entire domain (-∞, +∞), we need to ensure that the two pieces of the function match at the point of transition, which is x = 2.
For the function to be continuous at x = 2, the limit of f(x) as x approaches 2 from both sides (left and right) should be equal.
Let's calculate the left-hand limit (LHL):
lim(x→2-) (cx + 2)
Since x approaches 2 from the left side, we can consider x values that are slightly less than 2. Therefore, we can substitute x = 2 into the expression:
LHL = c(2) + 2 = 2c + 2
Now let's calculate the right-hand limit (RHL):
lim(x→2+) (cx^2 - 2)
Since x approaches 2 from the right side, we can consider x values that are slightly more than 2. Therefore, we can substitute x = 2 into the expression:
RHL = c(2)^2 - 2 = 4c - 2
For the function to be continuous at x = 2, LHL = RHL. Therefore, we have:
2c + 2 = 4c - 2
Now we can solve this equation for the value of c:
2 + 2 = 4c - 2
4 = 4c
c = 1
Therefore, the constant C that makes the function continuous on (-∞, +∞) is C = 1.