1. find the values of s for which each function is continous:
f(x) = [ x if x less than or equal to 1]
[2x-1 if x greater than 1]
2. determine all values of x at which the function is discontinuous:
f(x) = (x^2-2x)/(x^2-3x+2)
Could someone please tell me how to do these problems I'm lost.
1. f(x) as defined is continuous on (-∞,+∞). The only check you need to do to be sure is at x=1.
f(1-)=1
f(1+)=2*1-1=1
So f(x) is continuous also at x=1.
2.
For discontinuity of a rational function, look for values of x which render the denominator zero.
You can do this by factorization.
To find the values of s for which the function f(x) is continuous, we need to examine the two cases separately.
1. For x ≤ 1:
In this case, the function is defined as f(x) = x. To determine the continuity, we need to check if the limit of f(x) as x approaches 1 from the left (x < 1) is equal to the value of f(x) at x = 1.
So, we evaluate the limit:
lim(x→1-) f(x) = lim(x→1-) x = 1
Then, we check the value of f(x) at x = 1:
f(1) = 1
Since the limit and the value of the function at x = 1 are equal, the function f(x) is continuous for x ≤ 1.
2. For x > 1:
In this case, the function is defined as f(x) = 2x - 1. To determine the continuity, we need to check if the limit of f(x) as x approaches 1 from the right (x > 1) is equal to the value of f(x) at x = 1.
So, we evaluate the limit:
lim(x→1+) f(x) = lim(x→1+) (2x - 1) = 2(1) - 1 = 1
Then, we check the value of f(x) at x = 1:
f(1) = 2(1) - 1 = 1
Since the limit and the value of the function at x = 1 are equal, the function f(x) is continuous for x > 1.
The values of s for which the function f(x) is continuous are x ≤ 1 and x > 1.
Now, let's move on to the second question.
To determine the values of x at which the function f(x) = (x^2 - 2x) / (x^2 - 3x + 2) is discontinuous, we need to look for values of x that make the denominator of the function equal to zero.
The denominator x^2 - 3x + 2 can be factored as (x - 1)(x - 2).
Setting the denominator equal to zero gives us:
x - 1 = 0 => x = 1
x - 2 = 0 => x = 2
These are the values of x at which the denominator becomes zero, which means the function f(x) is discontinuous at x = 1 and x = 2.
Hence, the values of x at which the function f(x) is discontinuous are x = 1 and x = 2.