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.
Sure! Let's break down each problem step by step:
1. Finding the values of s for which the function is continuous:
To find the values of s for which the function is continuous, we need to check if the two parts of the function match up at the point where their switch occurs, which is x = 1.
First, let's consider the left-hand side of the function: f(x) = x for x ≤ 1.
This part of the function is defined for all values of x that are less than or equal to 1.
Next, let's consider the right-hand side of the function: f(x) = 2x-1 for x > 1.
This part of the function is defined for all values of x that are greater than 1.
To make the function continuous, we need the value of the function at x = 1 to be the same for both parts. So, we need to find the value of f(x) at x = 1 for both parts of the function.
For the left-hand side (x ≤ 1), f(1) = 1.
For the right-hand side (x > 1), f(1) = 2(1) - 1 = 1.
Since the value of f(x) at x = 1 is the same for both parts of the function, the function is continuous at x = 1.
Therefore, the function f(x) is continuous for all values of x.
2. Determining all values of x at which the function is discontinuous:
To find the values of x at which the function is discontinuous, we need to look for points where the function is not defined or where there is a vertical asymptote.
For the given function f(x) = (x^2-2x)/(x^2-3x+2), we need to find the values of x that make the denominator equal to zero. This is because division by zero is undefined.
To find the values of x that make the denominator zero, we solve the equation x^2-3x+2 = 0 by factoring or using the quadratic formula.
The factored form of the equation is (x-1)(x-2) = 0.
Setting each factor equal to zero, we find x = 1 and x = 2.
Therefore, the function is discontinuous at x = 1 and x = 2, as these values make the denominator zero.
So, the values of x at which the function is discontinuous are x = 1 and x = 2.
I hope this helps! Let me know if you have any further questions.