Please help I'm stuck on this problem for the past hour.
Define f(x)= { Ln lx-1l }/x
A) show that f(c) is continuous at x=2
B)Where on the interval [-2,2] is f discontinuous? Show the work that leads to your conclusion.
C) classify the discontinuities in part b) as removable or no removable.
Like, I know that IVT is involved, but how is it used to solve this?
well, when x=0 the function is undefined. When x=1, it is undefined because of the ln function.
f(0) is undefined, but the limit is -1 at x=0.
So, there's a removable discontinuity at x=0.
Naturally, at x=1, f is undefined; there's an asymptote there.
Take a look at the graph for further justification of your analysis:
http://www.wolframalpha.com/input/?i=ln%7Cx-1%7C%2Fx
To solve these problems, let's go step by step:
A) To show that f(c) is continuous at x = 2, we need to evaluate the left-hand limit, right-hand limit, and the function value at x = 2 and demonstrate that they all match.
To find the left-hand limit, we evaluate f(x) as x approaches 2 from the left side (x < 2).
Limit as x approaches 2 from the left of f(x) = Limit as x approaches 2 from the left [ln| x - 1 |] / x
To find the right-hand limit, we evaluate f(x) as x approaches 2 from the right side (x > 2).
Limit as x approaches 2 from the right of f(x) = Limit as x approaches 2 from the right [ln| x - 1 |] / x
Finally, we evaluate f(2) by directly substituting x = 2 into the function.
f(2) = [ln| 2 - 1 |] / 2
If the left-hand limit, right-hand limit, and the function value all match, then f(c) is continuous at x = 2.
B) To determine where f(x) is discontinuous on the interval [-2, 2], we look for values of x where the function fails to be continuous. There are two potential locations for discontinuities: at x = -1 and x = 1.
We need to examine the limit of f(x) as x approaches -1 and as x approaches 1 from both the left and right sides.
To find the limit as x approaches -1 from the left of f(x), we evaluate f(x) as x approaches -1 from the left side (x < -1).
Limit as x approaches -1 from the left of f(x) = Limit as x approaches -1 from the left [ln| x - 1 |] / x
Next, we find the limit as x approaches -1 from the right side (x > -1).
Limit as x approaches -1 from the right of f(x) = Limit as x approaches -1 from the right [ln| x - 1 |] / x
Looking at these limits, we observe that they are not equal on both sides of -1.
Similarly, we find the limits for x = 1:
Limit as x approaches 1 from the left of f(x) = Limit as x approaches 1 from the left [ln| x - 1 |] / x
Limit as x approaches 1 from the right of f(x) = Limit as x approaches 1 from the right [ln| x - 1 |] / x
Again, the limits are not equal on both sides of 1.
Therefore, the function f(x) is discontinuous at x = -1 and x = 1.
C) To classify the discontinuities from Part B as removable or non-removable, we need to examine if the function can be continuously modified at those points.
If the left-hand limit, right-hand limit, and the function value at the point are all defined and finite, we classify the discontinuity as removable. This means that we can "fill in the hole" and redefine the function at those points to make it continuous.
If any of the left-hand limit, right-hand limit, or function value at the point is undefined or infinite, we classify the discontinuity as non-removable. This means that no matter how we redefine the function at that point, it cannot be made continuous.
By evaluating the left-hand limit, right-hand limit, and the function value at x = -1 and x = 1, we can determine whether the discontinuities are removable or non-removable.