Question

Can anyone help me with these two problems?

Answer 1

sorry I can't seem to flip it!

Answer 2

11. E)
12. A)

Answer 3

Whoops, sorry. Forgot explanations...

11 is simple if you know the definition of a limit, there really isn't much to say there..

12 to do this problem it's good to visualize it. Draw the two points x = 0 and x = 2. Then draw the line y = 1/2. You'll see that for there to be two solutions (y = 1/2 intercepts f(x) twice) the point between x = 0 and x = 2, x = 1, must be below y = 1/2. y = 0 is the only point below.

Answer 4

1 is e, and I could provide examples of equations that fit all those if you would like. 2 is a because the Intermediate Value Theorem applies for the intervals [0,1] and [1,2].

Answer 5

Let's see if I can help:

For #1, lets consider what a limit means. If the limit as x approaches a of a function equals L, this means that the value of the function approaches L as you move toward point a from both the left and the right. However, the limit tells us nothing about the value of the function at a. Assuming the limit as x approaches a is L, f(a) could be L (f is continuous), f(a) could be another arbitrary number (f has a floating point discontinuity at a), or f(a) could be undefined (f has a hole at a). Thus we see that B, C, and D are all clearly not necessarily true. Furthermore, if f(a) is undefined, clearly f'(a) is undefined, so A cannot be correct. Thus, the correct answer is E.

For #2, we use the intermediate value theorem. This theorem tells us that if a continuous function takes on 2 y values f(a) and f(b) at x values a and b, it must take on every y value between f(a) and f(b) at some point between a and b. For example, if a continuous function takes on the value 1 and x=3 and 3 at x=5, then there is an x value in [3,5] for which f(x) is any number in the interval [1,3]. Thus, to show that the continuous function f takes on the value 1/2 at least twice, we want to show that the value 1/2 is straddled by 2 pairs of points. Thus, we choose k = 0, so that the function must pass through 1/2 to get from y=1 at x=0 to y=0 at x=1, and pass through it again to get from y=0 to y=2 at x=2.

Answer 6

thank you! what is the difference between a floating discontinuity and a hole?

Answer 7

A floating discontinuity is when the function is real, but not equal to the limit (mostly only piecewise) where a hole is when the function is not real (y = x^2/x)

Answer 8

A floating point discontinuity is when a function is continuous on both sides of a point and both sides move toward a given value, but the actual value of the function at that point is something else. They're usually seen in piecewise functions; for example, the function
\[f = x, x \neq 3, f(3) = 7\] has a floating point discontinuity at 3 because the function moves toward 3 from both sides of f(3), but the actual value is something else.

A hole is when a function is undefined at a point; for example, 1/x is undefined at x=0 and as such has a hole.

It is also worth noting that there is a third type of discontinuity - a jump discontinuity. This occurs when the function jumps from one value to another at a given point and is the only kind of discontinuity where the limit does not exist. An example of this would be \[f = 3, x \le 5, f = 4, x > 5\] because the function jumps from 3 to 4 at x = 5 with no continuity.

Answer 9

Right, right. When he said "hole" I was only thinking removable because that's what we referred to the removable discontinuities as in my class.

Answer 10

Oh yes, hole, removable discontinuity, and removable point discontinuity are all interchangeable terms.

Answer 11

I'm having some trouble understanding the IVT because 0 is not between 1 and 2