The function f is defined by f(x)= x^2-2x+3 if x≤1 and f(x)=-2x+5 if x > 1
Is f continuous at x=1? How do I solve this
let's find f(1) for both functions
if f(x) = x^2-2x+3 , then f(1) = 1 - 2 + 3 = 2
if f(x) = -2x + 5 , then f(1) = -2 + 5 = 3
so clearly there is a "jump" at x = 1
let's look at the graph
http://www.wolframalpha.com/input/?i=plot+f(x)%3D+x%5E2-2x%2B3+,+f(x)%3D-2x%2B5
delete the part of the parabola which is greater than 1
delete the part of the straight line, which is less than 1
can you see the final result, and the gap from 2 to 3 in the y value?
Thank you. How would I be able to tell if this is continuous without using a graph though?
To determine if a function is continuous at a specific point, we need to check if the left-hand limit, the right-hand limit, and the function value at that point are all equal.
In this case, we need to find the left-hand limit, the right-hand limit, and the function value at x=1.
First, let's find the left-hand limit:
lim(x→1-) f(x) = lim(x→1-) (x^2 - 2x + 3)
Since the function is defined as x^2 - 2x + 3 for x≤1, we can substitute x=1 into the function:
lim(x→1-) f(x) = (1^2 - 2(1) + 3) = 2
Next, let's find the right-hand limit:
lim(x→1+) f(x) = lim(x→1+) (-2x + 5)
Since the function is defined as -2x + 5 for x > 1, we substitute x=1:
lim(x→1+) f(x) = (-2(1) + 5) = 3
Now, let's find the function value at x=1:
f(1) = (1^2 - 2(1) + 3) = 2
To determine if f is continuous at x=1, we need to check if the left-hand limit, the right-hand limit, and the function value at x=1 are all equal. In this case, the left-hand limit is 2, the right-hand limit is 3, and the function value is 2. Since the left-hand limit is not equal to the right-hand limit, f is not continuous at x=1.