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?
2 answers
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
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