Suppose f(x)=3x(x-1)/x²-3x+2 for x≠1,2,f(1)= -3,f(2)=4 then f(x) is continuous
a) except at x=1
b) except at x=2
c) except at x =1 or 2
d) except at x= 0,1, or 2
e) at each real number
x²-3x+2 = (x-1) (x-2)
so your function is
f(x)=3x/(x-2) for x≠1,2,f(1)= -3,f(2)=4
yes f(1) = -3 , no problem with continuous there if f(1) given as -3
but at x = 2 the thing blows up as we approach x = 2 and saying that f(2)= 4 does not help.
To determine the continuity of f(x), we need to check if it is continuous at x = 1 and x = 2.
Let's start by checking the left-hand limit and right-hand limit at x = 1.
1. Left-hand limit:
Lim(x→1-) [f(x)] = Lim(x→1-) [3x(x-1)/(x²-3x+2)]
= Lim(x→1-) [3x(x-1)/[(x-1)(x-2)]]
= Lim(x→1-) [3x/(x-2)]
= 3/(1-2) = -3
2. Right-hand limit:
Lim(x→1+) [f(x)] = Lim(x→1+) [3x(x-1)/(x²-3x+2)]
= Lim(x→1+) [3x(x-1)/[(x-1)(x-2)]]
= Lim(x→1+) [3x/(x-2)]
= 3/(1-2) = -3
Since the left-hand limit (-3) is equal to the right-hand limit (-3), the limit exists and is equal to -3.
Next, let's check the value of f(x) at x = 1.
f(1) = -3, as given.
Therefore, since the left-hand limit, right-hand limit, and the value of f(x) at x = 1 are all equal (-3), f(x) is continuous at x = 1.
Now, let's move on and check the continuity at x = 2.
Similarly, we calculate the left-hand limit, right-hand limit, and the value of f(x) at x = 2.
1. Left-hand limit:
Lim(x→2-) [f(x)] = Lim(x→2-) [3x(x-1)/(x²-3x+2)]
= Lim(x→2-) [3x(x-1)/[(x-1)(x-2)]]
= Lim(x→2-) [3x/(x-1)]
= 6/1 = 6
2. Right-hand limit:
Lim(x→2+) [f(x)] = Lim(x→2+) [3x(x-1)/(x²-3x+2)]
= Lim(x→2+) [3x(x-1)/[(x-1)(x-2)]]
= Lim(x→2+) [3x/(x-1)]
= 6/1 = 6
f(2) = 4, as given.
However, the left-hand limit (6) is not equal to the right-hand limit (6). Therefore, the limit does not exist at x = 2.
Since the limit does not exist at x = 2, f(x) is not continuous at x = 2.
In conclusion, f(x) is continuous except at x = 2. Therefore, the correct answer is b) except at x = 2.
To determine if a function is continuous at a particular point, we need to check three conditions:
1. The function should be defined at that point.
2. The limit of the function as x approaches that point should exist.
3. The function value should equal the limit as x approaches that point.
Let's evaluate these conditions for the given function f(x)=3x(x-1)/(x²-3x+2).
Condition 1:
The function is defined for all real numbers except x=1 and x=2. However, we are given f(1)=-3 and f(2)=4, which means the function is also defined at these specific points.
Condition 2:
We need to evaluate the limit of f(x) as x approaches 1 and 2.
For x approaching 1:
lim(x→1) of f(x) = lim(x→1) of (3x(x-1)/(x²-3x+2))
To find this limit, we can substitute x=1 into the function: f(1) = -3.
Therefore, the limit of f(x) as x approaches 1 exists and equals -3.
For x approaching 2:
lim(x→2) of f(x) = lim(x→2) of (3x(x-1)/(x²-3x+2))
To find this limit, we can substitute x=2 into the function: f(2) = 4.
Therefore, the limit of f(x) as x approaches 2 exists and equals 4.
Condition 3:
Since f(1)=-3 and lim(x→1) of f(x)=-3, the function value is equal to the limit at x=1.
Similarly, since f(2)=4 and lim(x→2) of f(x)=4, the function value is equal to the limit at x=2.
Based on these evaluations, we can conclude that the function f(x) is continuous at each real number except at x=1 or 2. Therefore, the correct option is:
c) except at x=1 or 2.